Question

A solution containing $8.0\, g$ of nicotine in $92\, g$ of water freezes $0.925$ degrees below the normal freezing point of water. If the moles freezing point depression constant, $k_f = 1.85^{\circ}C\, mol^{-1}$, then the molar mass of nicotine is

• A. 16
• B. 80
• C. 320
• D. 160

Answer 1

Answer: (D)

Given,
Weight of nicotine, $w_2 = 8\, g$
Weight of water, $w_1 = 92\, g$
$\Delta T_{f}=0.925^{\circ}C$
$k_{f}=1.85^{\circ}C \,mol^{-1}$
The relation between molar mass and freezing point is given by
$M=\frac{k_{f}\times w_{2}\times1000}{\Delta T_{f}\times w_{1}}$
$=\frac{1.82\times8\times1000}{0.925\times92}=171\,g$
[Closest to $160\,g$]