Question

The dissolution of 5.25 g of a substance in 565 g of benzene at
298 K raises the boiling point by 0.625 K. Note that K_{f}
= 5.12 K kg/mol, K_{b} = 2.53 K kg/mol, and the density of
benzene is 876.6 kg/m^{3} .

Calculate the:

a.) freezing point depression

b.) ratio of the vapor pressure above the solution to that of the pure solvent

c.) the osmotic pressure

d.) molar mass of the solute.

Given: (P^{*}_{benzene} = 103 Torr at 298 K)

Answer #1

a)

dTb = Kb*m

m = 0.625 / 2.53

m = 0.24703 mol of solute / kg solvent

we have m = 565 g = 0.565 kg

mol of solute = 0.24703*0.565 =0.13957 mol of solute

a)

dTF = -K*m

dTf = -5.12 * 0.24703 = -1.26479 °C

Tf = 5.5 - 1.26479 = 4.23521°C

b)

vpaor pressure vs pure solvent

dP = xsolute * P°solvent

mol solvnet = mass/NMW = 565/78.11 = 7.2333 mol of B

xsolute = 0.13957 / (0.13957 + 7.2333) = 0.01893

Ratio of rpression

dP = 0.01893 *103 = 1.94979

Pmix = 103-1.94979 =101.05 torr

so..

Ratio = 101.05/103 = 0.9810

c)

P =MRT

M = mol/V

v = mass/D = (565)/(0.8766) = 644.53 mL = 0.64453 L

M = mol solute / Volume solution = 0.13957 /0.64453 = 0.21654 M

P = M*RT = 0.21654*0.082*298 = 5.2913 atm

d)

molar mass of solute

MW = mass / mol = 5.25/0.13957 = 37.615 g/mol

The dissolution of 5.25 g of a substance in 565 g of benzene at
298 K raises the boiling point by 0.625°C. Note that Kf = 5.12 K
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