Question

The molal freezing point constant for water is $1.86$. If $342g$ of cane sugar is dissolved in $1000g$ of water, the solution will freeze at:

• A. ${1.86}^{o}C$
• B. $-{1.86}^{o}C$
• C. $-{3.92}^{o}C$
• D. ${3.92}^{o}C$

Answer 1

Answer: (B)

$-{1.86}^{o}C$

$\Delta T_f={-k}_{f}m$

molality of solution = $\left(342/342\right)=1$

$\Delta T_f=-1.86\times \left(342/342\right)/1={-1.86}^{o}C$