Question

If $5g$ of $N{a}_{2}S{O}_4$ is dissolved in $45g$ of $H_{2}O$, the freezing point is changed by ${3.72}^{\circ }C$. Calculate the van't Hoff factor for $N{a}_{2}S{O}_4$.
Given: Freezing point depression constant for water is ${1.86}^{\circ }Ckgmo{l}^{-1}$

• A. 1.85
• B. 2.56
• C. 3.11
• D. 3.8

Answer 1

The correct option is B 2.56

van't Hoff factor is a correction factor by which actual extent of association/ dissociation can be expressed.
To correct the colligative property we multiply colligative property with 'i'.
The corrected colligative property of depression in freezing point is given by,
$△T_f=i\times K_f\times m$
where,
$△T_f$ is the depression in freezing point.
$K_f$ is molal depression constant.
m is molality of solution.
i is van't Hoff factor
$△T_f=i\times K_f\times \frac{w_B\times 1000}{m_B\times w_A}$
where,
$w_b$ is mass of solute
$m_B$ is molar mass of solute
$w_A$ is mass of solvent

Given:
$K_f={1.86}^{\circ }Ckgmo{l}^{-1}$
$△T_f={3.72}^{\circ }C$
$w_b=5g$
$m_B=142g/mol$
$w_A=45g$
$\therefore$
$3.72=i\times 1.86\times \frac{5\times 1000}{142\times 45}$
$i=2.56$