Question

The molar freezing point constant for water is $1.86^{\circ} C / mol$. If $342\, g$ of cane sugar $\left(C_{12} H_{22} O_{11}\right)$ 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, $\Delta T_{f}=K_{f} . m=1.86 \times 1 m=1.86^{o}$
Hence, freezing point of solution
$=0.00-(1.86)=-1.86^{\circ} C$