Question

A solution contains 3.75 g of a nonvolatile pure hydrocarbon in 95 g acetone. The boiling point of pure acetone is 55.95°C and of the solution is 56.50°C. The Kb for acetone is 1.71°C/m. What is the molar mass of the hydrocarbon?

Answer 1

Answer: The molar mass of the hydrocarbon is 37 g/mol

Explanation:

Elevation in boiling point is given by:

$\Delta T_b=i\times K_b\times m$

$\Delta T_b=T_b-T_b^0=(56.50-55.95)^0C=0.55^0C$ = Elevation in boiling point

i= vant hoff factor =1  (for non electrolyte )

$K_b$ = boiling point constant = $0.512^0C/m$

m= molality

$\Delta T_b=i\times K_b\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}$

Weight of solvent (acetone) = 95 g = 0.095 kg

Molar mass of solute = M g/mol

Mass of solute = 3.75 g

$0.55^0C=1\times 0.512\times \frac{3.75g}{Mg/mol\times 0.095kg}$

$M=37g/mol$

Thus the molar mass of the hydrocarbon is 37 g/mol