Chemistry · 2021
JEE Main · 27 August 2021, Shift 1 · Q53
1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to -4^° C before freezing. The amount of ice (in g ) that will be separated out is \…
$\displaystyle 1$ kg of $\displaystyle 0.75$ molal aqueous solution of sucrose can be cooled up to $\displaystyle -4^{\circ} \mathrm{C}$ before freezing. The amount of ice (in g ) that will be separated out is \_\_\_\_. (Nearest integer)
[Given : $\displaystyle \mathrm{K}_f\left(\mathrm{H}_2 \mathrm{O}\right)=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ ]
Official answer
518
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JEE Main 2021 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.