Verified Solution • 2025–26 Session moderate

QJEEM24Ch2f55: JEE Main 2024 Chemistry - Class 12 JEEM24Ch2f55 Solution: of each of water and acetic acid are mixed. The freezing point…

2.7 kg2.7 \mathrm{~kg} of each of water and acetic acid are mixed. The freezing point of the solution will be −x∘C-x^{\circ} \mathrm{C} . Consider the acetic acid does not dimerise in water, nor dissociates in water. x=x= ________ (nearest integer)

[Given: Molar mass of water =18 g mol−1=18 \mathrm{~g} \mathrm{~mol}^{-1} , acetic acid =60 g mol−1=60 \mathrm{~g} \mathrm{~mol}^{-1}

KfH2O:1.86 K kg mol−1{ }^{\mathrm{K}_{\mathrm{f}}} \mathrm{H}_2 \mathrm{O}: 1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}

Kf\mathrm{K}_{\mathrm{f}} acetic acid: 3.90 K kg mol−13.90 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}

freezing point: H2O=273 K\mathrm{H}_2 \mathrm{O}=273 \mathrm{~K} , acetic acid =290 K=290 \mathrm{~K} ]

Step-by-Step Verified Solution

amp; Molality of acetic acid =270060×12.7 mol/kgamp;=16.667amp;ΔTf=Kf×16.667amp;=1.86×16.667amp;=31 K\begin{aligned}& \text { Molality of acetic acid }=\frac{2700}{60} \times \frac{1}{2.7} \mathrm{~mol} / \mathrm{kg} \\&=16.667 \\& \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \times 16.667 \\&=1.86 \times 16.667 \\&=31 \mathrm{~K}\end{aligned}

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