Calculate vant Hoff Factor for Naphthoic Acid in Benzene

To determine the van't Hoff factor ($i$), we use the formula for the depression in freezing point of a solution:

$\Delta T_f = i \cdot K_f \cdot m$

where:

• $\Delta T_f$ is the freezing point depression $= 2\text{ K}$
• $K_f$ is the molal freezing point depression constant of the solvent $= 1.72\text{ K kg mol}^{-1}$
• $m$ is the molality of the solution in $\text{mol kg}^{-1}$
• $i$ is the van't Hoff factor

Step 1: Calculate the molar mass ($M$) of naphthoic acid ($\text{C}_{11}\text{H}_8\text{O}_2$)

$M = (11 \times 12) + (8 \times 1) + (2 \times 16)$
$M = 132 + 8 + 32 = 172\text{ g mol}^{-1}$

Step 2: Calculate the molality ($m$) of the solution

The number of moles of solute ($n$) is given by:
$n = \frac{\text{mass of solute}}{\text{molar mass}} = \frac{20\text{ g}}{172\text{ g mol}^{-1}}$

The mass of the solvent (benzene) in kilograms is:
$W_{\text{solvent}} = 50\text{ g} = 0.05\text{ kg}$

Thus, the molality ($m$) is:
$m = \frac{n}{W_{\text{solvent}}} = \frac{20 / 172}{0.05} = \frac{20}{172} \times \frac{1000}{50} = \frac{400}{172}\text{ mol kg}^{-1}$

Step 3: Calculate the van't Hoff factor ($i$)

Substitute the known values into the freezing point depression equation:

$2 = i \times 1.72 \times \left(\frac{400}{172}\right)$

Since $\frac{1.72}{172} = 0.01$, the equation simplifies to:

$2 = i \times 0.01 \times 400$
$2 = i \times 4$
$i = \frac{2}{4} = 0.5$

Therefore, the van't Hoff factor ($i$) is 0.5.

Correct Option: A