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  3. Chemistry
  4. At room temperature, a dilute solution of urea is prepared by dissolving 0.60 g of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mmHg, lowering of vapour pressure will be (molar mass of urea = 60 g mol-1):

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Q. At room temperature, a dilute solution of urea is prepared by dissolving $0.60\, g$ of urea in $360\, g$ of water. If the vapour pressure of pure water at this temperature is $35\, mmHg$, lowering of vapour pressure will be (molar mass of urea = $60\, g$ mol$^{-1}$):

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Solution:

Lowering of vapour pressure $ = p^0 - p = p^0 . x_{solute}$
$\therefore \Delta p =35 \times\frac{0.6/60}{\frac{0.6}{60} + \frac{360}{18}} $
$ = 35 \times\frac{0.1}{.01+20} = 35\times\frac{.01}{20.01} $
= $.017 mm\, Hg$