Question

The freezing point depression constant for water is $-1.86^{\circ}C\, m^{-1}$. If $6.00\, g\, Na_{2}SO_{4}$ is dissolved in $54.0\, g\, H_{2}O$, the freezing point is changed by $-3.82^{\circ}C$. Calculate the van’t Hoff factor for $Na_{2}SO_{4}$.

• A. $2.05$
• B. $2.63$
• C. $3.11$
• D. $0.381$

Answer 1

Answer: (B)

We know that $\Delta T_{f} = i \times K_{f} \times \frac{w_{B} \times 1000}{w_{B} \times w_{A}}$
Given: $\Delta T_{f} = 3.82, K_{f} = 1.86$
$w_{B} = 6\, m_{B} = 142, w_{A} = 54$
$\Delta T_{f} = i\times K_{f} \times \frac{w_{B} \times 1000}{m_{B} \times w_{A}}$
$i = \frac{\Delta T_{f} \times m_{B} \times w_{A}}{K_{f} \times w_{B} \times 1000}$
$ = \frac{3.82 \times 142 \times 54}{1.86 \times 6 \times 1000} = 2.63$