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Chemistry · 2021

JEE Main · 27 August 2021, Shift 2 · Q56

40 g of glucose (Molar mass =180 ) is mixed with 200 mL of water. The freezing point of solution is \ \ \ \ K. (Nearest integer) [Given: K_f=1.86 K…

$\displaystyle 40$ g of glucose (Molar mass $\displaystyle =180$ ) is mixed with $\displaystyle 200$ mL of water. The freezing point of solution is \_\_\_\_ K. (Nearest integer) [Given : $\displaystyle \mathrm{K}_f=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$; Density of water $\displaystyle =1.00 \mathrm{~g} \mathrm{~cm}^{-3}$; Freezing point of water $\displaystyle =273.15 \mathrm{~K}$ ]
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JEE Main 2021 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.