40 g of glucose (Molar mass)| LIQUID SOLUTION| JEE-MAIN 2021

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JEE MAIN 2021

27.08.21_S2

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40 g of glucose (Molar mass $=180$ ) is mixed with 200 mL of water. The freezing point of solution is $\_\_\_\_$ K. (Nearest integer) [Given : $\mathrm{K}_{\mathrm{f}}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$; Density of water $=1.00 \mathrm{~g} \mathrm{~cm}^{-3}$; Freezing point of water $=273.15 \mathrm{~K}$ ]

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