Question

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 $5.03^{\circ} C$ $\left(K_{f}\right.$ of benzene $\left.=5.12 \,K \,mol ^{-1} kg \right) .$ Molar mass of the compound is

• A. $162\, g \,mol ^{-1}$
• B. $-156\, g \,mol ^{-1}$
• C. $145 \, g \,mol ^{-1}$
• D. $156\, g \,mol ^{-1}$

Answer 1

Answer: (D)

On adding $\left(\frac{w_{1}}{m_{1}}\right)$ mole of solute to $w_{2} g$ of solvent, depression in freezing point is

$\Delta T_{f}=\frac{1000\, K_{f} w_{1}}{m_{1} w_{2}}$

$(15.51-5.03)=\frac{1000 \times 5.12 \times 0.643}{m_{1} \times(50 \times 0.879)}$

$m_{1}=156 \,g\, mol ^{-1}$