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JEE Main 20244 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2024 · 4 April · Evening Shift · Chemistry · Solutions

(2.7 ~kg ) of each of water and acetic acid are mixed. The freezing...

(2.7  kg\mathrm{~kg}) of each of water and acetic acid are mixed. The freezing point of the solution will be (-x^{{}^{\circ}} C\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\mathrm{~g} \mathrm{~mol}^{-1}), acetic acid (=60  g mol−1\mathrm{~g} \mathrm{~mol}^{-1}) (\mathrm{K}_{\mathrm{f}} \mathrm{H}}_{2}\mathrm{O}:1.86 K kg mol−1)(Kf)aceticacid:(3.90 K kg mol−1)freezingpoint:(H: 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} O=\mathrm{O}=273  K\mathrm{~K}), acetic acid (=290  K\mathrm{~K})]

Correct answer

0

Step-by-step solution

As moles of water >> moles of CH3COOH\mathrm{CH}_3 \mathrm{COOH} water is solvent. TF∘−(TF)S=KF×M0−(TF)S=1.86×2700/602700/1000( TF)S=−31∘C.\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}
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