Question

The molar freezing point constant for water is $ 1.86{}^\circ C/mole $ . If $342\, g$ of cane sugar $ (C_{12}H_{22}O_{11})$ is dissolved in $1000 \,g$ of water, the solution will freeze at:

• A. $ -1.86{}^\circ C $
• B. $ 1.86{}^\circ C $
• C. $ -3.92{}^\circ C $
• D. $ 2.42{}^\circ C $

Answer 1

Answer: (A)

Molality of cane sugar solution
$ =\frac{342}{342\times 1}=1\,m $
We know that $ \Delta {{T}_{f}}={{K}_{f}}.m $
$ =1.86\times 1 $
$ =1.86{}^\circ $
Hence, freezing point of solution
$ =0.00-(1.86)=-1.86{}^\circ C $