The freezing point depression constant for water is $1.86 K . K g^{- 1}$ If 5g of $N a_{2} S O_{4}$ is dissolved in 45g of $H_{2} O,$ the freezing point is changed by $- {3.82}^{0} C$ Calculate the van’t Hoff factor for $N a_{2} S O_{4}$

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3.11

0.381

2.05

2.63

answer is B.

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Detailed Solution

We know that, $Δ T_{f} = i \times K_{f} \times \frac{W_{B} \times 1000}{m_{B} \times w_{A}}$

Given

$\begin{array}{l} Δ T_{f} = 3.82, K_{f} = 1.86 \\ w_{B} = 5, m_{B} = 142, w_{A} = 45 \\ i = \frac{Δ T_{f} \times m_{B} \times w_{A}}{K_{f} \times w_{B} \times 1000} = \frac{3.82 \times 142 \times 45}{1.86 \times 5 \times 1000} = 2.63 \end{array}$

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