Question

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

• A. 16
• B. 80
• C. 320
• D. 160

Answer 1

Answer: (D)

Given,
Weight of nicotine, $w_2=8g$
Weight of water, $w_1=92g$
$\Delta T_f=0.925^{\circ }C$
$k_f=1.85^{\circ }Cmo{l}^{-1}$
The relation between molar mass and freezing point is given by
$M=\frac{k_f\times w_2\times 1000}{\Delta T_f\times w_1}$
$=\frac{1.82\times 8\times 1000}{0.925\times 92}=171g$
[Closest to $160g$]