JEE MAIN - Chemistry (2019 - 10th April Morning Slot - No. 9)
At room temperature, a dilute solution of urea is prepared by dissolving 0.60 of urea in 360 g of water. If the
vapour pressure of pure water at this temperature is 35 mm Hg, lowering of vapour pressure will be.
(molar mass of urea = 60 g mol–1)
0.031 mmHg
0.017 mmHg
0.028 mmHg
0.027 mmHg
Explanation
Given that,
wsolute = wurea = 0.6 gm
wsolvent = wH2O = 360 gm
po = 35
We know,
lowering of vapour pressure
$$\Delta $$p = xsolute $$ \times $$ po
= $${{{n_{urea}}} \over {{n_{urea}} + {n_{{H_2}O}}}}$$ $$ \times $$ po
= $${{{{0.6} \over {60}}} \over {{{0.6} \over {60}} + {{360} \over {18}}}}$$ $$ \times $$ 35
= $${{{{10}^{ - 2}}} \over {20}} \times 35$$
= 0.017
wsolute = wurea = 0.6 gm
wsolvent = wH2O = 360 gm
po = 35
We know,
lowering of vapour pressure
$$\Delta $$p = xsolute $$ \times $$ po
= $${{{n_{urea}}} \over {{n_{urea}} + {n_{{H_2}O}}}}$$ $$ \times $$ po
= $${{{{0.6} \over {60}}} \over {{{0.6} \over {60}} + {{360} \over {18}}}}$$ $$ \times $$ 35
= $${{{{10}^{ - 2}}} \over {20}} \times 35$$
= 0.017
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