$A$ solution containing $8.0 \, g$ of nicotine in $92 \, g$ of water freezes $0.925 \, ^{\circ}C$ below the normal freezing point of water. If the molal freezing point depression constant,$k_f = 1.85 \, ^{\circ}C \, kg \, mol^{-1}$,then the molar mass of nicotine is $...$
- A$16 \, g \, mol^{-1}$
- B$80 \, g \, mol^{-1}$
- C$320 \, g \, mol^{-1}$
- D$160 \, g \, mol^{-1}$
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