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Solutions question

2024 · 4 Apr · Shift 2 · Q25
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Solutions question

2024 · 4 Apr · Shift 2 · Q25

JEE MainChemistrySolutionsNumerical+4 / −1
2.7 kg2.7 \mathrm{~kg}2.7 kg of each of water and acetic acid are mixed. The freezing point of the solution will be −x∘C-x^{\circ} \mathrm{C}−x∘C. Consider the acetic acid does not dimerise in water, nor dissociates in water. x=x=x=‾\underline{\hspace{2cm}}​ (nearest integer) [Given: Molar mass of water =18 g mol−1=18 \mathrm{~g} \mathrm{~mol}^{-1}=18 g mol−1, acetic acid =60 g mol−1=60 \mathrm{~g} \mathrm{~mol}^{-1}=60 g 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​H2​O:1.86 K kg mol−1 Kf\mathrm{K}_{\mathrm{f}}Kf​ acetic acid: 3.90 K kg mol−13.90 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}3.90 K kg mol−1 freezing point: H2O=273 K\mathrm{H}_2 \mathrm{O}=273 \mathrm{~K}H2​O=273 K, acetic acid =290 K=290 \mathrm{~K}=290 K]
Numerical answer
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