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)

$-1.86{}^\circ C$

Answer 2

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 $=0.00-(1.86)=-1.86{}^\circ C$