Question

3 attempts left Check my work Be sure to answer all parts A solution of 5.06 g of a compound having the empirical formula CoHjP in 25.0 g of benzene is observed to freeze at 2.3°C. Calculate the molar mass of the solute and its molecular formula. 615 Molar Mass Molecular Formula The normal freezing point of benzene is 5.50°C and the Kf for benzene is 5.12°C/m.

Answer 1

a)
Δ Tf = 5.50 - 2.3 = 3.2 oC

use:
Δ Tf = Kf*mb
3.2 = 5.12 *mb
mb= 0.625 molal


m(solvent)= 25.0 g
= 0.025 Kg

number of mol,
n = Molality * mass of solvent in Kg
= (0.625 mol/Kg)*(0.025 Kg)
= 1.562*10^-2 mol

mass(solute)= 5.06 g


use:
number of mol = mass / molar mass
1.562*10^-2 mol = (5.06 g)/molar mass
molar mass = 324 g/mol
Answer: 324 g/mol

b)

Molar mass of C6H5P,
MM = 6*MM(C) + 5*MM(H) + 1*MM(P)
= 6*12.01 + 5*1.008 + 1*30.97
= 108.07 g/mol

Now we have:
Molar mass = 324.0 g/mol
Empirical formula mass = 108.07 g/mol
Multiplying factor = molar mass / empirical formula mass
= 324.0/108.07
= 3

Hence the molecular formula is : C18H15P3