vapor pressure of solutions and freezing-point depression/boiling-point elevation problems?

Question

VAPOR PRESSURE OF SOLUTIONS
A. 10.0 g sample of the solid sucrose ( sugar, C12H22O11) was dissolved in 200.0 g of water. What will the vapor pressure of this solution be at 25C, if the normal vapor pressure of pure water at 25C is 23.8 torr?

B. Predict the vapor pressure of a solution prepared by dissolving 49.4 g of solid calcium chloride, CaCl, in150 g of water at 25C. The vapor pressure of water at 25C is 23.8 torr.


FREEZING POINT DEPRESSION/BOILING POINT ELEVATION OF SOLUTIONS
A. The freezing point of pure liquid benzene, C6H6, is 5.48 C. when 20.0 g of an unknown compound is dissolved in 500.0 g of benzene, the freezing poioint of the resulting solution is 3.77 C. what is the molecular weight of the unknown compount if Kf / (benzene) = 5.12 C. kg/mol?

B. What is the normal boiling point of a solution prepared by dissolving 54.6 grams of potassium chloride, KCl, in 400 mL of water (d = 0.997 g/mL, Kb= 0.51 C kg/mol)?

step by step process please. thanks

Answer 1

Vapor Pressure of Solutions

A.

Amount of the sugar present = 10.0g/ (342) = 0.0292397661 mol
Amount of water present = 200g/ (18) = 11.11111111 mol
Thus, X(H2O), which signifies the fraction of water present,
= 11.11111111/ (11.11111111 + 0.0292397661)
= 0.9973753281
At 25 deg Celsius,
Vapor Pressure of Water is then = 0.9973753281 x 23.8
= 23.73753281 Torr.


B.

Amount of Calcium Chloride = 49.4g/ (40.1) = 1.2319202 mol
Amount of water = 150g/ (18) = 8.333333333 mol
Thus, X(H2O), which signifies the fraction of water present,
= 8.333333333/ (8.333333333 + 1.2319202 )
= 0.8712088294
Vapor Pressure of Water is then = 0.8712088294 x 23.8
= 20.73477014 Torr.


FREEZING POINT DEPRESSION/BOILING POINT ELEVATION OF SOLUTIONS

A - NOTE: T(f) signifies T with the small f at its bottom right. Similarly, b(f) is an 'f' with a small f at its bottom right.

dT(f) = change in freezing point

Change in dT(f) = 5.48 - 3.77 = 1.71
Therefore, from {c}
b(f), which signifies the molality,
= 1.71/ 5.12 = 0.333984375

Since molality = (Amount of solute added)/ (mass of solvent/ kg),

Amount of solute added = [500 x 10^(-3)]x(0.333984375)
= 0.1669921875mol

Thus, Mr of the unknown compound is 20.0/ (0.1669921875)
= 119.7660819

B

Change in T(b) = (0.51) x {54.6/ [0.997x400x10^(-3)]}
= 69.82447342
since dT(f) = K x b(f) and
molality = (Amount of solute added)/ (mass of solvent/ kg).

I am afraid I do not know how to proceed from here for I am not sure of the method to find the normal boiling point of a solution.