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  4. 60 g of urea is dissolved in 1100 g solution. To keep Δ Tf/Kf as 1 mol kg-1 , water separated in the form of ice is

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Q. 60 g of urea is dissolved in 1100 g solution. To keep $ \Delta {{T}_{f}}/{{K}_{f}} $ as 1 mol $ k{{g}^{-1}} $ , water separated in the form of ice is

MGIMS WardhaMGIMS Wardha 2013

Solution:

1100 g solution has 60 g urea Water =1040 g $ \Delta {{T}_{f}}=\frac{1000{{K}_{f}}{{w}_{2}}}{{{M}_{2}}{{w}_{1}}} $ $ \frac{\Delta {{T}_{f}}}{{{K}_{f}}}=\frac{1000\times 60}{60\times {{w}_{1}}}=1 $ or $ {{w}_{1}}=1000\,g $ The ice formed $ =1040-1000=40\text{ }g $