Question

What would be the freezing point of a solution that has $684$ grams of sucrose ($1$ mol $=342$g) dissolved in $2,000$ grams of water (1 mole= 18 grams)?

• A. $-{1.86}^o$C or $271.14$K
• B. $-{0.93}^o$C or $272.07$K
• C. $-{1.39}^o$C or $271.61$K
• D. $-{2.48}^o$C or $270.52$K
• E. $-{3.72}^o$C or $269.28$K

Answer 1

The correct option is A $-{1.86}^o$C or $271.14$K

Number of moles of sucrose $=\frac{mass}{molarmass}=\frac{684}{342}=2$ moles
Molality of sucrose solution, $m=\frac{\text{number of moles}}{\text{volume of water in kg}}=\frac{2}{2}=1$ m
Note : 2000 g water = 2 kg.
The depression in the freezing point, $\Delta T_f=K_{f}m=1.86\times 1=1.86$
The freezing point of water is 0 deg C. The freezing point of solution will be $0-1.86=-{1.86}^{o}C=273-1.86K=271.14K$.