\(2.7 \mathrm{~kg}\) of each of water and acetic acid are mixed. The freezing point of the solution will be…
\(2.7 \mathrm{~kg}\) of each of water and acetic acid are mixed. The freezing point of the solution will be \(-x^{\circ} \mathrm{C}\). Consider the acetic acid does not dimerise in water, nor dissociates in water. \(x=\) ______ (nearest integer)
[Given: Molar mass of water \(=18 \mathrm{~g} \mathrm{~mol}^{-1}\), acetic acid \(=60 \mathrm{~g} \mathrm{~mol}^{-1}\)
\(\mathrm{K}_{\mathrm{f}} \mathrm{H}_2 \mathrm{O}: 1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
\(\mathrm{K}_{\mathrm{f}}\) acetic acid: \(3.90 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\)
freezing point: \(\mathrm{H}_2 \mathrm{O}=273 \mathrm{~K}\), acetic acid \(=290 \mathrm{~K}\)]
Solution
As moles of water $>$ moles of $\mathrm{CH}_3 \mathrm{COOH}$ water is solvent.
$\begin{aligned}
& \mathrm{T}_{\mathrm{F}}^{\circ}-\left(\mathrm{T}_{\mathrm{F}}\right)_{\mathrm{S}}=\mathrm{K}_{\mathrm{F}} \times \mathrm{M} \\
& 0-\left(\mathrm{T}_{\mathrm{F}}\right)_{\mathrm{S}}=1.86 \times \frac{2700 / 60}{2700 / 1000} \\
& \left(\mathrm{~T}_{\mathrm{F}}\right)_{\mathrm{S}}=-31^{\circ} \mathrm{C} .
\end{aligned}$