$1.2 \mathrm{~mL}$ of acetic acid having density $1.06 \mathrm{~g} \mathrm{~cm}^{-3}$ is dissolved in 1 litre of water. The depression in freezing point observed for this concentration of acid was $0.041^{\circ} \mathrm{C}$. The van't Hoff factor of the acid is $\left(K_f\right.$ of water $\left…

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$1.2 \mathrm{~mL}$ of acetic acid having density $1.06 \mathrm{~g} \mathrm{~cm}^{-3}$ is dissolved in 1 litre of water. The depression in freezing point observed for this concentration of acid was $0.041^{\circ} \mathrm{C}$. The van't Hoff factor of the acid is $\left(K_f\right.$ of water $\left.=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$

Aicharya · Accepted Answer

From depression in freezing point,

$ \begin{aligned} & \text { Molality }(m)=\frac{\text { Moles of solute }}{\text { Mass of solvent }(\text { in } \mathrm{g})} \times 1000 \\ & \text { Mass }=\text { density } \times \text { volume }=1.2 \times 1.06=1.272 \mathrm{~g} \\ & \therefore \text { Moles of solute }=\frac{1.272}{60} \\ & \qquad\left(\because \text { Moles }=\frac{\text { Mass }}{\text { Molecular mass }}\right) \\ & 0.041=\frac{i \times 1.86 \times 1.272>\times 1000}{60 \times 1000}=\frac{60 \times 0.041}{1.86 \times 1.27} \\ & \quad i=1.04 \end{aligned} $ Thus, option (b) is correct

$1.2 \mathrm{~mL}$ of acetic acid having density $1.06 \mathrm{~g} \mathrm{~cm}^{-3}$ is dissolved in 1…

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