Introduction to general chemistry. The molar shares of the substance and water in 1000 g of water at 20 dissolves

Example 1. Calculate the osmotic pressure of the solution containing in 1.5 l 135 g of glucose C 6 H 12 O 6 at 0 0 C.

Decision:Osmotic pressure is determined by the law of Vant-Gooff:

SM RT.

The molar concentration of the solution is found by the formula:

Substituting the value of the molar concentration into the expression of the Law of the Vant-Gooff, we calculate the osmotic pressure:

π \u003d S M RT\u003d 0.5 mol / l ∙ 8,314 Pa ∙ m 3 / mol ∙ K ∙ 273 \u003d 1134,86 ∙ 10 3

Example 2.Determine the boiling point of the solution containing 1.84 g of nitrobenzene C 6 H 5 NO 2 in 10 g of benzene. The boiling point of pure benzene 80.2 0 C.

Decision: The boiling point of the solution on Δt instrumentation will be higher than the boiling point of pure benzene: t of instrumentation (solution) \u003d t of instrumentation (solvent) + Δt kip;

By the law of Raoul: Δt kip \u003d e ∙ with m ,

where E. -Ebulloscopic solvent constant (table value),

With M. - molaous concentration of solution, mole / kg

Δt kip \u003d e ∙ with m \u003d1.5 ∙ 2.53 \u003d 3.8 0 S.

t Kip (solution) \u003d T Kip (solvent) + Δt kip \u003d80.2 0 C +3.8 0 C \u003d 84 0 C.

901. A solution containing 57 g of sugar from 12 H 22 o 11 in 500 g of water, boils at 100.72 0 C. Determine the ebulloscopic water constant.

902. A solution containing 4.6 g of glycerin C 3 H 8 O 3 in 71 g of acetone, boils at 56.73 0 C. Determine the ebulloscopic constant of acetone if the boiling point of acetone 56 0 C.

903. Calculate the boiling point of the solution containing 2 g of naphthalene from 10 H 8 to 20 g of ether if the boiling point of the ether is 35.6 0 s, and its ebulloscopic constant 2,16.

904. 4 g of substances are dissolved in 100 g of water. The resulting solution freezes at -0.93 0 C. Determine the molecular weight of the dissolved substance.

905. Determine the relative molecular weight of benzoic acid, if a 10% solution boils at 37.57 0 C. The boiling point of the ester is 35.6 0 s, and its ebulloscopic constant 2,16.

906. Lowering the freezing temperature of a solution containing 12.3 g of nitrobenzene C 6 H 5 NO 2 to 500 g of benzene is 1.02 0 C. Determine the cryoscopic benzene constant.

907. The freezing temperature of acetic acid 17 0 s, a cryoscopic constant 3.9. Determine the temperature of the freezing of the solution containing 0.1 mol of the dissolved substance in 500 g of acetic acid CH 3 of the coxy.

908. The solution containing 2.175 g of the dissolved substance in 56.25 g of water freezes at -1.2 0 C. Determine the relative molecular weight of the dissolved substance.

909. At what temperature, a solution brings a solution containing 90 g of glucose from 6 H 12 o 6 in 1000 g of water?

910. In 200 g of alcohol dissolved 5 g of substance. The solution is piling at 79.2 0 C. Determine the relative molecular weight of the substance, if the Ebulloscopic constant of alcohol 1.22. Boiling point of alcohol 78.3 0 S.

911. The aqueous sugar solution freezes at -1.1 0 C. Determine the mass proportion (%) of sugar from 12 H 22 o 11 in solution.

912. In which mass of water, 46 g of glycerin C 3 H 8 O 3 should be dissolved to obtain a solution with a boiling point of 100,104 0 s?

913. The solution containing 27 g of substance is 1 kg of water, boils at 100.078 0 C. Determine the relative molecular weight of the dissolved substance.

914. Calculate the mass of water in which 300 g of glycerol with 3 H 8 O 3 should be dissolved to obtain a solution freezing at - 2 0 C.

915. The glucose solution in water shows an increase in the boiling point by 0.416 0 C. Employ the decrease in the freezing temperature of this solution.

916. Calculate the freezing temperature of 20% glycerol solution with 3 H 8 O 3 in water.

917. In 250 g of water dissolved 1.6 g of substance. The solution freezes at -0.2 0 C. Calculate the relative molecular weight of the dissolved substance.

918. The solution containing 0.5 g of acetone (CH 3) 2 CO in 100 g of acetic acid gives a decrease in the freezing temperature by 0.34 0 C. Determine the cryoscopic constant of acetic acid.

919. Calculate the mass proportion (%) of glycerol in aqueous solution, the boiling point of which is 100.39 0 C.

920. How many grams of ethylene glycol C 2 H 4 (OH) 2 is required to add to each kilogram of water to prepare antifreeze with a freezing point -9.3 0 s?

921. A solution containing 565 g of acetone and 11.5 g of glycerol with 3 H 5 (OH) 3 boils at 56.38 0 C. Pure acetone boils 56 0 C. Calculate the ebulloscopic constant of acetone.

922. At what temperature freezes a 4% solution of ethyl alcohol with 2N 5 in water?

923. Determine the mass fraction (%) of sugar from 12 H 22 O 11 in an aqueous solution if the solution boils at 101.04 0 C.

924. Which of the solutions will freeze at a lower temperature: 10% glucose solution with 6 H 12 O 6 or 10% Sugar solution from 12 H 22 o 11?

925. Calculate the freezing temperature of 12% aqueous (by weight) of the glycerin solution with 3 H 8 O 3.

926. Calculate the boiling point of the solution containing 100 g of sucrose C 12 H 22 O 11 in 750 g of water.

927. The solution containing 8.535 g of Nano 3 in 100 g of water is crystallized at T \u003d -2.8 0 C. Determine the cryoscopic constant water.

928. To prepare a coolant on 20 liters of water, 6 g of glycerol (\u003d 1.26 g / ml) was taken. What will be the freezing temperature of the cooked antifreeze?

929. Determine the amount of ethylene glycol C 2 H 4 (OH) 2, which must be added to 1 kg of water to prepare a solution with a crystallization temperature -15 0 C.

930. Determine the crystallization temperature of a solution containing 54 g of glucose C 6 H 12 O 6 in 250 g of water.

931. The solution containing 80 g of naphthalene C 10 H 8 in 200 g of diethyl ether, boils at T \u003d 37.5 0 C, and a pure ester - at T \u003d 35 0 C. Determine the ebuloscopic ether constant.

932. When adding 3.24 g of sulfur in 40 g of benzene C 6 H 6, the boiling point was increased by 0.91 0 C. Solo particles consist of a sulfur particle in solution if the ebulloscopic benzene constant is 2.57 0 C.

933. The solution containing 3.04 g of camphor C 10 H 16 O in 100 g of benzene with 6 H 6, boils at T \u003d 80,714 0 C. (The boiling point of benzene is 80.20 0 s). Determine the Eculoscopic constant of benzene.

934. How many grams of carbamide (urea) CO (NH 2) 2 must be dissolved in 125 g of water so that the boiling point is increased by 0.26 0 C. Ecobulloscopic water constant 0.52 0 C.

935. Calculate the boiling point of 6% (by weight) of the aqueous solution of glycerol C 3 H 8 O 3.

936. Calculate the mass proportion of sucrose from 12 H 22 o 11 in an aqueous solution, the crystallization temperature of which is 0.41 0 C.

937. When dissolving 0.4 g of some substance in 10 g of water, the crystallization temperature of the solution dropped to 1.24 0 C. Calculate the molar mass of the dissolved substance.

938. Calculate the temperature of the freezing of 5% (by weight) of the sugar solution from 12 H 22 O 11 in water.

939. How many grams of glucose C 6 H 12 o 6 should be dissolved in 300 g of water to obtain a solution with a boiling point of 100, 5 0 s?

940. The solution containing 8.5 g of some non-electrolyte in 400 g of water, boils at a temperature of 100.78 0 C. Calculate the molar mass of the dissolved substance.

941. When dissolved 0.4 g of some substance in 10 g of water, the crystallization temperature of the solution was -1.24 0 C. Determine the molar mass of the dissolved substance.

942. Calculate the mass fraction of sugar from 12 H 22 O 11 in the solution, the boiling point of which is 100, 13 0 C.

943. Calculate the crystallization temperature of 25% (by weight) of the glycerol solution with 3 H 8 O 3 in water.

944. The crystallization temperature of benzene with 6 H 6 5.5 0 s, the cryoscopic constant is 5.12. Calculate the molar mass of nitrobenzene, if a solution containing 6.15 g of nitrobenzene in 400 g of benzene crystallizes at 4.86 0 C.

945. Glycerol solution C 3 H 8 O 3 in water shows an increase in boiling point by 0.5 0 C. Calculate the crystallization temperature of this solution.

946. Calculate the mass fraction of urea (NH 2) 2 in aqueous solution, the crystallization temperature of which is5 0 S.

947. In which water should be dissolved 300 g of benzene with 6 H 6 to obtain a solution with a crystallization temperature -20 0 s?

948. Calculate the boiling point of 15% (by weight) of the glycerol solution with 3 H 8 O 3 in acetone, if the boiling point of acetone 56.1 0 s, and the ebuloscopic constant is 1.73.

949. Calculate the osmotic pressure of the solution at 17 0 s, if it contains 18.4 g of glycerol C 3 H 5 (OH) 3.

950. In 1 ml of the solution contains 10 15 molecules of the dissolved substance. Calculate the osmotic pressure of the solution at 0 0 C. In which volume contains 1 mol of the dissolved substance?

951. How many molecules of the dissolved substance are contained in 1 ml of the solution, the osmotic pressure of which at 54 0 s is 6065 Pa?

952. Calculate the osmotic pressure of 25% (by weight) of the sucrose solution C 12 H 22 O 11 at 15 0 C (ρ \u003d 1.105 g / ml).

953. At what temperature, the osmotic pressure of the solution containing in 1 l of water 45 g of glucose C 6 H 12 O 6 will reach 607.8 kPa?

954. Calculate the osmotic pressure of 0.25m sugar solution with 12 H 22 O 11 at 38 0 C.

955. At what temperature, the osmotic pressure of the solution containing in 1 l 60 g of glucose from 6 H 12 o 6 will reach 3 atm?

956. Osmotic pressure of the solution, the volume of which is 5 liters, at 27 0 s is 1.2 ∙ 10 5 Pa. What is the molar concentration of this solution?

957. How many grams of ethyl alcohol C 2 H 5 should contain 1 liters of solution so that its osmotic pressure is the same as a solution containing in 1 l at the same temperature of 4.5 g of formaldehyde CH 2 O.

958. How many grams of ethyl alcohol C 2 H 5 should it be dissolved in 500 ml of water so that the osmotic pressure of this solution at 20 0 s is 4.052 ∙ 10 5 pa?

959. 200 ml of solution contains 1 g of dissolved substance and at 20 0 s have an osmotic pressure of 0.43 ∙ 10 5 Pa. Determine the molar mass of the dissolved substance.

960. Determine the molar mass of the dissolved substance, if a solution containing in 0.5 l 6 g of substance, at 17 0 s, has a osmotic pressure of 4.82 ∙ 10 5 Pa.

961. How many grams of glucose C 6 H 12 O 6 should contain 1 liters of solution so that its osmotic pressure is the same as a solution containing in 1 l at the same temperature of 34.2 g of sugar C 12 H 22 O 11?

962. 400 ml of solution contain 2 g of dissolved substance at 27 0 C. Osmotic pressure of the solution 1,216 ∙ 10 5 Pa. Determine the molar mass of the dissolved substance.

963. Sugar solution C 12 H 22 O 11 at 0 0 C has an osmotic pressure of 7.1 ∙ 10 5 Pa. How many grams of sugar are contained in 250 ml of such a solution?

964. 2.45 g of carbamide is contained in 7 liters of solution. Osmotic pressure of the solution at 0 0 s is 1.317 ∙ 10 5 Pa. Calculate the molar mass of the carbamide.

965. Determine the osmotic pressure of the solution, in 1 l of which contains 3.01 ∙ 10 23 molecules at 0 0 C.

966. The aqueous solutions of phenol with 6 H 5 It and glucose C 6 H 12 O 6 contain in 1 l equal masses of dissolved substances. In which of the solutions, the osmotic pressure is greater at the same temperature? How many times?

967. The solution containing 3 g of non-electrolyte in 250 ml of water freezes at temperatures - 0.348 0 C. Calculate the molar mass of non-electrolyte.

968. The solution containing in 1 l 7.4 g of glucose C 6 H 12 O 6 at a temperature of 27 0 ° C has the same osmotic pressure with the hydrogen solution CO (NH 2) 2. How many urea is contained in 500 ml of solution?

969. Osmotic pressure of the solution, in 1 l of which contains 4.65 g of aniline C 6 H 5 NH 2, at a temperature of 21 0 C is 122.2 kPa. Calculate the molar mass of aniline.

970. Calculate osmotic pressure at a temperature of 20 0 С 4% sugar solution with 12 H 22 O 11, the density of which is 1.014 g / ml.

971. Determine the osmotic pressure of the solution containing 90.08 g of glucose C 6 H 12 O 6 in 4 liters at a temperature of 27 0 C.

972. The solution, a volume of 4 liters, contains at a temperature of 0 0 ° C 36.8 g of glycerin (C 3 H 8 O 3). What is the osmotic pressure of this solution?

973. At 0 0, the osmotic pressure of the sucrose solution from 12 H 22 o 11 is 3.55 ∙ 10 5 Pa. What is the mass of sucrose contains 1 liter of solution?

974. Determine the magnitude of the osmotic solution, in 1 liter of which from0.4 mol of non-electrolyte at a temperature of 17 0 C.

975. What is equal to the osmotic pressure of a solution containing in 2.5 liters of a solution of 6.2 g of aniline (C 6 H 5 NH 2) at a temperature of 21 0 C.

976. At 0 0, the osmotic pressure of the sucrose solution with 12 H 22 o 11 is 3.55 ∙ 10 5 Pa. What is the mass of sucrose contains 1 liter of solution?

977. At what temperature will the an aqueous solution of ethyl alcohol be freezing, if the mass fraction of 2 H 5 is equal to 25%?

978. The solution containing 0.162 g of sulfur in 20 g of benzene boils at a temperature of 0.081 0 with higher than pure benzene. Calculate the molecular weight of sulfur in the solution. How many atoms contain in the same sulfur molecule?

979. To 100 ml of 0.5 mol / l of aqueous solution of sucrose from 12 N 22 o 11 Added 300 ml of water. What is equal to the osmotic pressure of the resulting solution at 25 0 s?

980. Determine the boiling and freezing temperature of a solution containing 1 g of nitrobenzene with 6 H 5 NO 2 in 10 g of benzene. Ebyloscopic and cryoscopic benzene constant is respectively 2.57 and 5.1 K ∙ kg / mol. Boiling point of pure benzene 80.2 0 C, freezing temperature -5.4 0 C.

981. What is the freezing temperature of a non-electrolyte solution containing 3.01 ∙ 10 23 molecules in one water liter?

982. Camphore solutions weighing 0.522 g in 17 g of ether boils at a temperature of 0.461 0 from higher than clean air. Ebulloscopic Empire constant 2,16 K ∙ kg / mol. Determine the molecular weight of the camphor.

983. The boiling point of the aqueous solution of sucrose is 101.4 0 C. Calculate the molaous concentration and mass fraction of sucrose in solution. At what temperature freezes this solution?

984. The molecular weight of the non-electrolyte is 123.11 g / mol. What is the mass of non-electrolyte must be contained in 1 liter of solution so that the solution at 20 0 s had an osmotic pressure equal to 4.56 ∙ 10 5 pa?

985. When dissolved 13.0 non-electrolyte in 400 g of diethyl ether (C 2 H 5) 2, the boiling point was increased by 0.453 K.. Determine the molecular weight of the dissolved substance.

986. Determine the boiling point of the water solution of glucose, if the mass fraction of 6 H 12 O 6 is 20% (for water to e \u003d 0.516 K ∙ kg / mol).

987. The solution consisting of 9.2 g of iodine and 100 g of methyl alcohol (CH 3) boils at 65.0 0 C. How many atoms are part of the iodine molecule located in a dissolved state? The boiling point of the alcohol is 64.7 0 s, and its ebulloscopic constant to e \u003d 0.84.

988. How many grams of sucrose C 12 H 22 o 11 should be dissolved in 100 g of water to: a) lower the crystallization temperature by 1 0 s; b) increase the boiling point of 1 0 s?

989. In 60 g of benzene dissolved 2.09 of some substance. The solution is crystallized at 4.25 0 C. Install the molecular weight of the substance. Clean benzene crystallizes at 5.5 0 C. Cososcopic benzene constant 5.12 K ∙ kg / mol.

990. At 20 0, the osmotic pressure of the solution, in 100 ml of which contains 6.33 g of the coloring agent of blood - hematine, is 243.4 kPa. Determine the molecular weight of hematine.

991. The solution consisting of 9.2 g of glycerol with 3 H 5 (O) 3 and 400 g of acetone boils at 56.38 0 C. Pure acetone boils at 56.0 0 C. Calculate the ebulloscopic acetone constant.

992. Water steam pressure at 30 0 s is 4245.2 pa. What mass of sugar from 12 H 22 o 11 should be dissolved in 800 g of water to obtain a solution, the pressure of the steam of which is 33.3 pa less water steam pressure? Calculate the mass proportion (%) of sugar in solution.

993. Ether pair pressure at 30 0 s is 8.64 ∙ 10 4 Pa. What amount of non-electrolyte should be dissolved in 50 mol ether to reduce the pressure of steam at a given temperature by 2666 pa?

994. Lowering steam pressure over a solution containing 0.4 mol aniline in 3.04 kg of servo carbon, at a certain temperature is 1003.7 Pa. The pressure of the survo carbon pair at the same temperature is 1.0133 ∙ 10 5 PA. Calculate the molecular weight of the carbon.

995. At some temperature, the steam pressure above the solution containing 62 g of phenol C 6 H 5 O to 60 mol ether is 0.507 ∙ 10 5 Pa. Find the ether steam pressure at this temperature.

996. Pressure steam of water at 50 0 s is 12334 Pa. Calculate the pressure pair pressure containing 50 g of ethylene glycol with 2 H 4 (O) 2 in 900 g of water.

997. Water vapor pressure at 65 0 s is 25003 pa. Determine the pressure of water vapor over a solution containing 34.2 g of sugar from 12 H 22 O 12 in 90 g of water at the same temperature.

998. Water steam pressure at 10 0 s is 1227.8 pa. In what volume of water should we dissolve 16 g of methyl alcohol to obtain a solution, the pressure of the steam of which is 1200 pa at the same temperature? Calculate the mass fraction of alcohol in solution (%).

999. At what temperature, an aqueous solution will crystallize, in which the mass fraction of methyl alcohol is 45%.

1000. A water-alcohol solution containing 15% alcohol is crystallized at - 10.26 0 C. Determine the molar mass of alcohol.

The properties of dilute solutions depending only on the amount of non-taxed solute substance are called convigative properties. These include lowering the pressure of the solvent steam over the solution, an increase in the boiling point and a decrease in the freezing temperature of the solution, as well as osmotic pressure.

Reducing the freezing temperature and an increase in the boiling point of the solution compared with a clean solvent:

T. Deputy. \u003d \u003d K. TO. m. 2 ,

T. kip. \u003d. = K. E. m. 2 .

where m. 2 - mortality of the solution, K. To I. K. E - Cyoscopic and Eculicopic constant solvent, X. 2 - the molar fraction of the dissolved substance, H. pl. and H. Span. - Enthalpy of melting and evaporation of the solvent, T. pl. and T. kip. - melting and boiling points of the solvent, M. 1 - molar mass of solvent.

Osmotic pressure in dilute solutions can be calculated by equation

where X. 2 - the molar fraction of the dissolved substance is the molar volume of the solvent. In very dilute solutions, this equation is converted to vant-Gooff equation:

where C. - molariness of the solution.

Equations describing the collegiate properties of non-electrolytees can also be applied to describe the properties of electrolyte solutions by entering the correction coefficient of Vant-Gooff i., eg:

= iCRT.or T. Deputy. \u003d. iK. TO. m. 2 .

Isotonic coefficient is associated with the degree of electrolyte dissociation:

i \u003d 1 + (- 1),

where - the number of ions formed during the dissociation of one molecule.

Solubility of solid in perfect solution at temperatures T. Describes schröder equation:

,

where X. - the molar fraction of the dissolved substance in solution, T. pl. - melting point and H. pl. - Enthalpy melting solved substance.

Examples

Example 8-1. Calculate the solubility of bismuth in the cadmium at 150 and 200 ° C. The enthalpy of bismuth melting at a melting point (273 o C) is 10.5 kJ. Mol -1. It is necessary that the perfect solution is formed and the melting enthalpy does not depend on temperature.

Decision. We use formula .

At 150 o C From! X. = 0.510

At 200 o C From! X. = 0.700

Solubility increases with temperature, which is characteristic of the endothermic process.

Example 8-2. A solution of 20 g of hemoglobin in 1 liter of water has an osmotic pressure of 7.52 10 -3 atm at 25 o C. Determine the molar mass of hemoglobin.

65 kg. Mol -1.

TASKS

1. Calculate the minimum osmotic work performed by the kidneys to isolate the urea at 36.6 o C, if the concentration of the urea in the plasma is 0.005 mol. L -1, and in the urine 0.333 mol. L -1.
2. 10 g of polystyrene was dissolved in 1 liter of benzene. The height of the decision of the solution (density of 0.88 G. cm -3) in osmometer at 25 ° C is equal to 11.6 cm. Calculate the molar mass of polystyrene.
3. Protein serum albumin man has a molar mass of 69 kg. Mol -1. Calculate the osmotic pressure of a solution of 2 g of protein in 100 cm 3 of water at 25 o C per PA and in MM of a solution of a solution. Consider the density of the solution equal to 1.0 cm -3.
4. At 30 ° C, the pressure of the aqueous solution of sucrose is 31.207 mm Hg. Art. Pressure a pair of clean water at 30 ° C is equal to 31.824 mm Hg. Art. The density of the solution is 0.99564. SM -3. What is the osmotic pressure of this solution?
5. The plasma of human blood freezes at -0.56 o C. What is its osmotic pressure at 37 o C, measured using a membrane, permeable for water only?
6. * The molar mass of the enzyme was determined, dissolving it in water and measuring the height of the solution of the solution in the osmometer at 20 o C, and then extrapolating the data to zero concentration. The following data obtained:

C.mg. SM -3.
h., cm

7. The molar mass of lipid is determined to increase the boiling point. Lipid can be dissolved in methanol or in chloroform. The boiling point of methanol 64.7 o C, the heat of evaporation of 262.8 cal. M -1. Boiling temperature Chloroform 61.5 o C, heat evaporation 59.0 cal. M -1. Calculate Eculicopic constant methanol and chloroform. What solvent is better to use to determine the molar mass with maximum accuracy?
8. Calculate the temperature of the freezing of aqueous solution containing 50.0 g of ethylene glycol in 500 g of water.
9. A solution containing 0.217 g of sulfur and 19.18 g CS 2, boils at 319.304 K. The boiling point of pure CS 2 is 319.2 K. Eculoscopic constant CS 2 is 2.37 K. kg. Mol -1. How many sulfur atoms are contained in sulfur molecule dissolved in CS 2?
10. 68.4 g of sucrose dissolved in 1000 g of water. Calculate: a) steam pressure, b) osmotic pressure, c) freezing temperature, d) boiling point of the solution. Pressure pressure of pure water at 20 ° C. equals 2314.9 Pa. Cyoscopic and eculoscopic permanent waters are equal to 1.86 and 0.52 K. kg. Mol -1, respectively.
11. The solution containing 0.81 g of hydrocarbon H (CH 2) N h and 190 g of ethyl bromide, freezes at 9.47 o C. The freezing temperature of the ethyl bromide 10.00 o C, the cryoscopic constant 12.5 K. kg. Mol -1. Calculate n.
12. When dissolving 1.4511 g of dichloroacetic acid in 56.87 g of carbon four-chloride, the boiling point rises by 0.518 degrees. Boiling point CCL 4 76.75 o C, heat evaporation 46.5 cal. M -1. What is the apparent molar mass of acid? What explains the discrepancy with the true molar mass?
13. A certain amount of substance dissolved in 100 g of benzene lowers the point of its freezing by 1.28 o C. The same amount of substance dissolved in 100 g of water lowers the point of its freezing by 1.395 o C. The substance has a normal molar mass in benzene, and in water Completely dissociated. How many ions the substance dissociates in aqueous solution? Cyoscopic constants for benzene and water are equal to 5.12 and 1.86 K. kg. Mol -1.
14. Calculate the perfect solubility of anthracene in benzene at 25 o C in units of poles. Anthracene melting enthalpy at melting point (217 o C) is 28.8 kJ. Mol -1.
15. Calculate solubility p- DibromBenzene in benzene at 20 and 40 o C, believing that the perfect solution is formed. Entalpy Melting p-dibrombeenzene at its melting temperature (86.9 o C) is equal to 13.22 kJ. Mol -1.
16. Calculate the solubility of naphthalene in benzene at 25 o C, believing that the perfect solution is formed. Enhaulpia melting naphthalene at a temperature of its melting (80.0 o C) is equal to 19.29 kJ. Mol -1.
17. Calculate the solubility of anthracene in toluene at 25 o C, believing that the perfect solution is formed. Anthracene melting enthalpy at melting point (217 o C) is 28.8 kJ. Mol -1.
18. Calculate the temperature at which pure cadmium is in equilibrium with a solution of Cd - Bi, the molar fraction of Cd in which is 0.846. Cadmium melting enthalpy at melting point (321.1 o C) is equal to 6.23 kJ. Mol -1.

Task 427..
Calculate the mogenic shares of alcohol and water in 96% (by mass) with a solution of ethyl alcohol.
Decision:
Moled share (N i) - the ratio of the amount of dissolved substance (or solvent) to the amount of the quantities of all
substances in solution. In a system consisting of alcohol and water, the molar proportion of water (N 1) is equal to

And the molar share of alcohol where N 1 is the number of alcohol; N 2 - the amount of water.

We calculate the mass of alcohol and water contained in 1 liter of the solution, provided that their density is equal to one from the proportions:

a) the mass of alcohol:

b) water weight:

We find the amount of substances by the formula:, where m (c) and m (c) - the mass and amount of substance.

Now we calculate the molar shares of substances:

Answer: 0,904; 0,096.

Task 428.
In 1 kg of water dissolved 666g con The density of the solution is 1.395 g / ml. Find: a) Mass stake Kon; b) molarity; c) praying; d) mole alkali and water shares.
Decision:
but) Mass fraction - the percentage of the mass of the dissolved substance to the total mass of the solution is determined by the formula:

where

m (p-ra) \u003d M (H 2 O) + M (KOH) \u003d 1000 + 666 \u003d 1666

b) the molar (volume-molar) concentration shows the number of moles of the dissolved substance contained in 1 liter of the solution.

We will find a mass of a horse, per 100 ml of solution by the formula: Formula: M \u003d p.V, where p is the density of the solution, V is the volume of the solution.

m (KOH) \u003d 1,395 . 1000 \u003d 1395

Now we calculate the molarity of the solution:

We find how many grams of HNO 3 accounted for 1000g water, reaching the proportion:

d) a molar fraction (N i) is the ratio of the amount of dissolved substance (or solvent) to the sum of the amounts of all substances in the solution. In a system consisting of alcohol and water, the molar proportion of water (N 1) is equal to a molar proportion of alcohol, where N 1 is the amount of the pitch. N 2 - the amount of water.

In 100g of this solution, 40g Kon 60g H2O is contained.

Answer: a) 40%; b) 9.95 mol / l; c) 11.88 mol / kg; d) 0.176; 0.824.

Task 429.
The density of 15% (by weight) solution H 2 SO 4 is 1.105 g / ml. Calculate: a) normal; b) molarity; c) praying solution.
Decision:
Find a mass of solution by the formula: M \u003d p. V, where p. - The density of the solution, V is the volume of the solution.

m (H 2 SO 4) \u003d 1,105 . 1000 \u003d 1105

Mass H 2 SO 4 contained in 1000 ml of solution, we find from the proportion:

We define the molar weight of the equivalent H 2 SO 4 from the relation:

M e (c) - the molar mass of the equivalent of acid, g / mol; M (c) - a molar mass of acid; Z (c) - equivalent number; Z (acid) is equal to the number of H + ions in H 2 SO 4 → 2.

a) The molar concentration of the equivalent (or normality) shows the number of equivalents of the dissolved substance contained in 1 liter of the solution.

b) Molant concentration

Now we calculate the praying of the solution:

c) Molaum concentration (or praying) shows the number of moles of the dissolved substance contained in 1000g solvent.

We find how many grams of H 2 SO 4 are contained in 1000g water, reaching the proportion:

Now we calculate the praying of the solution:

Answer: a) 3,38H; b) 1.69 mol / l; 1.80 mol / kg.

Task 430.
The density of the 9% (by weight) of the sucrose solution with 12 H 22 O 11 is 1.035 g / ml. Calculate: a) the concentration of sucrose in g / l; b) molarity; c) praying solution.
Decision:
M (from 12 H 22 o 11) \u003d 342g / mol. We will find a mass of solution according to the formula: m \u003d p v, where p is the density of the solution, V is the volume of the solution.

m (C 12 H 22 O 11) \u003d 1.035. 1000 \u003d 1035

a) Mass with 12 H 22 o 11 contained in solution, calculated by the formula:

where
- mass fraction of a dissolved substance; m (V-BA) - mass of dissolved substance; M (p-ra) - mass of solution.

The concentration of the substance in g / l shows the number of grams (vehicle units) contained in 1 l solution. Consequently, the concentration of sucrose is 93.15g / l.

b) The molar (volume-molar) concentration (C m) shows the number of moles of the dissolved substance contained in 1 liter of the solution.

in) Molant concentration (or molandom) shows the number of moles of the dissolved substance contained in 1000g solvent.

We find how many grams from 12 H 22 o 11 contains in 1000 g of water, reaching the proportion:

Now we calculate the praying of the solution:

Answer: a) 93.15 g / l; b) 0.27 mol / l; c) 0.29 mol / kg.

2.10.1. Calculation of the relative and absolute masses of atoms and molecules

The relative masses of atoms and molecules are determined using the Table of D.I. Mendeleev values \u200b\u200bof atomic masses. At the same time, when calculating the calculations for the study purposes, the values \u200b\u200bof atomic masses of elements are usually rounded to integer numbers (with the exception of chlorine, the atomic mass of which is taken equal to 35.5).

Example 1. Relative atomic mass of calcium and R (Ca) \u003d 40; Relative atomic mass of platinum A R (Pt) \u003d 195.

The relative mass of the molecule is calculated as the sum of the relative atomic masses of the components of the molecule of atoms, taking into account the number of their substance.

Example 2. Relative molar mass of sulfuric acid:

M R (H 2 SO 4) \u003d 2A R (H) + A R (S) + 4a R (O) \u003d 2 · 1 + 32 + 4· 16 = 98.

The values \u200b\u200bof absolute masses of atoms and molecules are divided by weight of 1 mol of matter for the number of Avogadro.

Example 3. Determine the mass of one calcium atom.

Decision. The atomic mass of calcium is R (Ca) \u003d 40 g / mol. The mass of one calcium atom will be equal to:

m (Ca) \u003d A R (CA): N a \u003d 40: 6.02 · 10 23 = 6,64· 10 -23

Example 4. Determine the mass of one sulfuric acid molecule.

Decision. The molar mass of sulfuric acid is equal to M R (H 2 SO 4) \u003d 98. Mass of one molecule M (H 2 SO 4) is equal to:

m (H 2 SO 4) \u003d M R (H 2 SO 4): N a \u003d 98: 6,02 · 10 23 = 16,28· 10 -23

2.10.2. Calculation of the amount of substance and calculation of the number of atomic and molecular particles according to known mass values \u200b\u200band volume

The amount of substance is determined by dividing its mass, expressed in grams, on its atomic (molar) mass. The amount of substance located in a gaseous state with N.U. is dividing its volume to a volume of 1 mol of gas (22.4 liters).

Example 5. Determine the amount of sodium substance N (Na) located in 57.5 g of metallic sodium.

Decision. The relative atomic mass of sodium is equal to R (Na) \u003d 23. The amount of substance we find the division of the mass of metallic sodium on its atomic mass:

n (Na) \u003d 57.5: 23 \u003d 2.5 mol.

Example 6. Determine the amount of nitrogen substance, if its volume at N.U. It is 5.6 liters.

Decision. Number of nitrogen N (n 2) we find the division of its volume on the volume of 1 mol of gas (22.4 liters):

n (n 2) \u003d 5.6: 22.4 \u003d 0.25 mol.

The number of atoms and molecules in the substance is determined by multiplying the amount of the substance of atoms and molecules to the number of Avogadro.

Example 7. Determine the number of molecules contained in 1 kg of water.

Decision. The amount of water substance we find the division of its mass (1000 g) on \u200b\u200bthe molar mass (18 g / mol):

n (H 2 O) \u003d 1000: 18 \u003d 55.5 mol.

The number of molecules in 1000 g of water will be:

N (H 2 O) \u003d 55.5 · 6,02· 10 23 = 3,34· 10 24 .

Example 8. Determine the number of atoms contained in 1 l (N.U.) oxygen.

Decision. The amount of substance oxygen, the volume of which under normal conditions is 1 l.

n (O 2) \u003d 1: 22,4 \u003d 4,46 · 10 -2 mol.

The number of oxygen molecules in 1 l (N.U.) will be:

N (O 2) \u003d 4,46 · 10 -2 · 6,02· 10 23 = 2,69· 10 22 .

It should be noted that 26.9 · 10 22 molecules will be contained in 1 l of any gas at N.U. Since the oxygen molecule is ductoman, the number of oxygen atoms in 1 liter will be 2 times more, i.e. 5.38. · 10 22 .

2.10.3. Calculation of the middle molar mass of the gas mixture and volumetric share
Gas contained in it

The average molar weight of the gas mixture is calculated on the basis of molar masses of the components of this mixture of gases and their volume fractions.

Example 9. Believing that the content (in bulk percentage) of nitrogen, oxygen and argon in the air is respectively 78, 21 and 1, calculate the average molar mass of air.

Decision.

M Rid \u003d 0.78 · M R (n 2) +0.21 · M R (O 2) +0.01 · M R (AR) \u003d 0.78 · 28+0,21· 32+0,01· 40 = 21,84+6,72+0,40=28,96

Or approximately 29 g / mol.

Example 10. The gas mixture contains 12 l NH 3, 5 L N 2 and 3 L H 2, measured at N.U. Calculate the volumetric shares of gases in this mixture and its average molar mass.

Decision. The total volume of the mixture of gases is V \u003d 12 + 5 + 3 \u003d 20 liters. Volumetric fractions of j gases will be equal:

φ (NH 3) \u003d 12: 20 \u003d 0.6; φ (n 2) \u003d 5: 20 \u003d 0.25; φ (H 2) \u003d 3: 20 \u003d 0.15.

The average molar mass is calculated on the basis of the volume fractions of the components of this mixture of gases and their molecular weights:

M \u003d 0.6 · M (NH 3) +0.25 · M (N 2) +0.15 · M (H 2) \u003d 0.6 · 17+0,25· 28+0,15· 2 = 17,5.

2.10.4. Calculation of the mass fraction of the chemical element in the chemical connection

The mass fraction of the chemical element is defined as the ratio of the mass of the atom of this element contained in this mass of the substance to the mass of this substance M. Mass fraction is a dimensionless value. It is expressed in fractions from one:

ω (x) \u003d m (x) / m (0<ω< 1);

or percentage

ω (x),% \u003d 100 m (x) / m (0%<ω<100%),

where ω (x) is the mass fraction of the chemical element x; m (x) - the mass of the chemical element x; M - Mass of the substance.

Example 11. Calculate the mass fraction of manganese in manganese oxide (VII).

Decision. Molar masses of substances are: M (Mn) \u003d 55 g / mol, m (o) \u003d 16 g / mol, M (Mn 2 O 7) \u003d 2M (Mn) + 7m (o) \u003d 222 g / mol. Consequently, the mass of Mn 2 O 7 amount of substance 1 mol is:

m (Mn 2 O 7) \u003d M (Mn 2 O 7) · n (Mn 2 O 7) \u003d 222 · 1 \u003d 222

From formula Mn 2 O 7 it follows that the amount of substance of manganese atoms is twice as much as the amount of the agent oxide substance (VII). It means

n (Mn) \u003d 2n (Mn 2 O 7) \u003d 2 mol,

m (Mn) \u003d n (Mn) · M (Mn) \u003d 2 · 55 \u003d 110 g

Thus, the mass fraction of manganese in manganese oxide (VII) is equal to:

ω (x) \u003d m (Mn): M (Mn 2 O 7) \u003d 110: 222 \u003d 0.495 or 49.5%.

2.10.5. The establishment of the formula of the chemical compound by its elemental composition

The simplest chemical formula of the substance is determined on the basis of the known magnitude of the mass fractions of the elements that are part of this substance.

Suppose there is a sample of the substance Na x P y o z mass M o. Consider how its chemical formula is determined if the amount of substance of the elements of the elements, their masses or mass fractions in a known mass of the substance are known. The formula of the substance is determined by the ratio:

x: y: z \u003d n (na): n (p): n (o).

This relationship will not change if each member is divided into the number of Avogadro:

x: y: z \u003d n (na) / n a: n (p) / n a: n (o) / n a \u003d ν (na): ν (p): ν (o).

Thus, to find the formula of the substance, it is necessary to know the relationship between the amounts of the substances of atoms in the same mass of the substance:

x: y: z \u003d m (na) / m r (na): m (p) / m r (p): m (o) / m R (o).

If you divide each member of the last equation for the mass of the sample M o, then we obtain an expression that allows you to determine the composition of the substance:

x: y: z \u003d ω (na) / m R (na): ω (p) / m r (p): ω (o) / m R (o).

Example 12. The substance contains 85.71 wt. % carbon and 14,29 masses. % hydrogen. Its molar mass is 28 g / mol. Determine the simplest and true chemical formulas of this substance.

Decision. The ratio between the number of atoms in the molecule with X H y is determined by dividing the mass fractions of each element on its atomic mass:

x: y \u003d 85.71 / 12: 14,29 / 1 \u003d 7.14: 14,29 \u003d 1: 2.

Thus, the simplest formula of substance - CH 2. The simplest formula of the substance does not always coincide with its true formula. In this case, the CH formula 2 does not correspond to the valence of the hydrogen atom. To find a true chemical formula, you need to know the molar mass of this substance. In this example, the molar mass of the substance is 28 g / mol. Separating 28 by 14 (the sum of atomic masses corresponding to the CH 2 formular unit), we obtain the true ratio between the number of atoms in the molecule:

We obtain the true formula of substance: from 2 H 4 - ethylene.

Instead of the molar mass for gaseous substances and vapors, a density of any gas or by air can be indicated on the problem of the problem.

In the case under consideration, the gas density through the air is 0.9655. Based on this magnitude, the molar mass of the gas can be found:

M \u003d m · D Rest \u003d 29 · 0,9655 = 28.

In this expression, M is the molar weight of the gas with x n y, m is the average molar mass of the air, D is the density of the gas from the X N by air. The resulting magnitude of the molar mass is used to determine the true formula of the substance.

In the task condition, the mass fraction of one of the elements may not be indicated. It is subtraction from the unit (100%) of the mass fractions of all other elements.

Example 13. Organic compound contains 38.71 mass. % carbon, 51.61 masses. % oxygen and 9.68 masses. % hydrogen. Determine the true formula of this substance, if the density of its oxygen vapors is 1.9375.

Decision. Calculate the relationship between the number of atoms in the molecule with X H y o z:

x: y: z \u003d 38.71 / 12: 9,68 / 1: 51.61 / 16 \u003d 3,226: 9,68: 3,226 \u003d 1: 3: 1.

The molar mass of M substance is:

M \u003d M (O 2) · D (O 2) \u003d 32 · 1,9375 = 62.

The simplest formula of substance CH 3 O. The sum of atomic masses for this formula unit will be 12 + 3 + 16 \u003d 31. We divide 62 to 31 and we obtain the true ratio between the number of atoms in the molecule:

x: y: z \u003d 2: 6: 2.

Thus, the true formula of substance with 2 H 6 o 2. This formula corresponds to the composition of the diatomic alcohol - ethylene glycol: CH 2 (OH) -CH 2 (OH).

2.10.6. Determination of the molar mass of matter

The molar mass of the substance can be determined based on the magnitude of the density of its vapor of gas with a known magnitude of the molar mass.

Example 14. The density of the vapor of some organic compound according to oxygen is equal to 1,8125. Determine the molar mass of this compound.

Decision. The molar mass of the unknown substance M x is equal to the product of the relative density of this substance D to the molar mass of the M substance, according to which the relative density value is determined:

M x \u003d D · M \u003d 1,8125. · 32 = 58,0.

Substances with the found value of the molar mass can be acetone, propionic aldehyde and allyl alcohol.

The molar weight of the gas can be calculated using its molar volume with N.U.

Example 15. Weight of 5.6 liters of gas at N.U. It is 5.046 g. Calculate the molar mass of this gas.

Decision. The molar volume of gas at N.P. is equal to 22.4 liters. Consequently, the molar mass of the desired gas is equal

M \u003d 5,046. · 22,4/5,6 = 20,18.

The desired gas is neon ne.

The Klapeyrone Mendeleev equation is used to calculate the molar mass of the gas, the volume of which is set under conditions different from normal.

Example 16. At a temperature of 40 o C and a pressure of 200 kPa, the mass of 3.0 liters of gas is 6.0 g. Determine the molar mass of this gas.

Decision. Substituting the known values \u200b\u200bin the Klapaireron-Mendeleev equation we get:

M \u003d MRT / PV \u003d 6.0 · 8,31· 313/(200· 3,0)= 26,0.

The gas under consideration is acetylene C 2 H 2.

Example 17. With a combustion of 5.6 liters (N.U.) of hydrocarbons, 44.0 g of carbon dioxide and 22.5 g of water was obtained. The relative density of hydrocarbon oxygen is equal to 1,8125. Determine the true chemical formula of hydrocarbon.

Decision. The hydrocarbon combustion reaction equation can be represented as follows:

With X H y + 0.5 (2x + 0.5y) o 2 \u003d x 2 + 0,5UN H 2 O.

The amount of hydrocarbon is 5.6: 22.4 \u003d 0.25 mol. As a result of the reaction, 1 mol of carbon dioxide and 1.25 mol of water is formed, which contains 2.5 mol of hydrogen atoms. When burning hydrocarbon with the amount of substance 1 mol, 4 mol of carbon dioxide and 5 mol of water is obtained. Thus, 1 mole of hydrocarbon contains 4 mole of carbon atoms and 10 mol of hydrogen atoms, i.e. Chemical hydrocarbon formula with 4 H 10. The molar mass of this hydrocarbon is equal to m \u003d 4 · 12 + 10 \u003d 58. Its relative oxygen density d \u003d 58: 32 \u003d 1,8125 corresponds to the magnitude given in the problem of the problem, which confirms the correctness of the found chemical formula.