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At room temperature, a dilute solution o...

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 mm Hg, lowering of vapour pressure will be `1.74 xx 10^(-x)` mm Hg. The numerical value of x is _____________. (Molar mass of urea = 60 g `mol^(-1)`).

Text Solution

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The correct Answer is:
b
`P_(0) to` Vapour pressure of pure solvent
`P_(s) to` Vapour pressure of pure solutioin.
Number of moles of urea `=(0.60)/60 = 0.01` mol
Number of moles of water `=(360)/18 = 20` mol
`P_(0) - P_(S) rArr` Lowering in Vapour - pressure
We know `(P_(0)-P_(S))/(P_(0)) = X_("solute")`
`P_(0)-P_(s) =(n_("solute"))/(n_("solute") + n_("solvent")) xx P_(0) =0.01/(0.01 + 20) xx 35 = (0.01)/(20.01) xx 35 =0.0174`
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