Question

What is the freezing point for the solution of $2,000$ grams of water (essentially $2L$) and $684$ grams of sucrose(molecular mass = $342g/mol$) dissolved in it?

• A. $-{1.86}^{\circ }C$
• B. $-{6.78}^{\circ }C$
• C. $-{3.72}^{\circ }C$
• D. $-{51}^{\circ }C$

Answer 1

Answer: (A)

$-{1.86}^{\circ }C$

The molar mass of sucrose is $342g/mol$.
$684g$ of sucrose corresponds to $\frac{684}{342}=2$ moles.
$2000g$ of water corresponds to $2kg$.
The molality of the solution is the number of moles of sucrose present in $2kg$ of solution.
$m=\frac{2mol}{2kg}=1m$
The depression in the freezing point constant for water is ${1.86}^{0}C/m$.
The depression in the freezing point,

$\Delta T_f=m\times K_f$

$=1\times 1.86$
$={1.86}^{o}C$.
Freezing point of water is $0^{o}C$.

The freezing point of solution will be $0^{o}C-\left({1.86}^{o}C\right)=-{1.86}^{o}C$.