Question

What is the molar mass of a solution of 5.00 g of a compound in 25.00 g of carbon tetrachloride (bp 76.8 °c; kb = 5.02 °c/m) that boils at 81.5 °c at 1 atm? (a) outline the steps necessary to answer the question?

Answer 1

(84.4C - 76.5C) / (5.03 C/m) = 1.5706 m

(1.5706 mol) / (1000 g CC14) X (25.00 G CC14) = 0.039265 mol

(5.00 g) / (0.039265 mol)  = 127 g/mol

Answer 2

Answer: 214 g/mol

Explanation:

Formula used for Elevation in boiling point :

$\Delta T_b=k_b\times m$

or,

$T_{solution}-T_{solvent}=k_b\times \frac{\text{Mass of solute}}{\text {Molar mass of solute}\times \text{ Mass of solvent in Kg}}$

where,

$T_b$ = change in boiling point  =$(81.5-76.8)^0C=4.7^0C$

$k_b$ = boiling point constant  =$5.02^0C/m$

m = molality  = $\frac{5g}{\text{ Molar mass of solute}}\times 0.025kg$

$4.7=5.02\times \frac{5g}{\text {Molar mass of solute}\times 0.025kg}$

${\text {Molar mass of solute}}=214g/mol$

Thus molar mass of solute is 214 g/mol.