Question

The addition of $0.643 \,g$ of a compound to $50 \, mL$ of benzene (density $=0.879 \, g \,mL ^{-1}$ ) lowers the freezing point from $5.51^{\circ} C$ to $5.03^{\circ} C$. If the freezing point constant, $K_{f}$ for benzene is $5.12 \,K \, kg \,mol ^{-1}$, the molar mass of the compound is approximately

• A. $156\, g\, mol^{-1}$
• B. $88\, g\, mol^{-1}$
• C. $60\, g\, mol^{-1}$
• D. $312\, g\, mol^{-1}$

Answer 1

Answer: (A)

Given, weight of compound,
$w_{2}=0.643\,g$
$k_{f}=5.12\,K \,kg\, mol^{-1}$
$\Delta T_{f}=T^{\circ}_{f}-T_{f}$
$278.51-278.03=0.48 \,K$
Volume of benzene $= 50\, mL$
Density of benzene $= 0.879\, g\, mL^{-1}$
$\therefore $ Weight of benzene,
$w_{1} = 50\times0.879 = 43.95\, g$
Molar mass of a compound can be calculated by using the following formula
$M_{2}=\frac{k_{f}\times w_{2}\times1000}{\Delta T_{f}\times w_{1}}$
$=\frac{5.12\times0.643\times1000}{0.48\times43.95}$
$=156.05\, g\, mol^{-1}$