At room temperature,a dilute solution of urea | vedclass

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(D) The lowering of vapour pressure is given by the formula: $\Delta p = p^o \cdot X_{solute}$Given:$p^o = 35 \ mm \ Hg$$w_{urea} = 0.60 \ g$,$M_{urea

Vedclass · Accepted Answer

$0.017$

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(D) The lowering of vapour pressure is given by the formula: $\Delta p = p^o \cdot X_{solute}$Given:$p^o = 35 \ mm \ Hg$$w_{urea} = 0.60 \ g$,$M_{urea} = 60 \ g \ mol^{-1}$$w_{water} = 360 \ g$,$M_{water} = 18 \ g \ mol^{-1}$Calculate moles of solute $(n)$ and solvent $(N)$:$n = \frac{0.60}{60} = 0.01 \ mol$$N = \frac{360}{18} = 20 \ mol$Calculate mole fraction of solute $(X_{solute})$:$X_{solute} = \frac{n}{n + N} = \frac{0.01}{0.01 + 20} = \frac{0.01}{20.01} \approx 0.00049975$Calculate lowering of vapour pressure $(\Delta p)$:$\Delta p = 35 	imes 0.00049975 \approx 0.01749 \ mm \ Hg$Rounding to the nearest value,we get $0.017 \ mm \ Hg$.

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 $35 \ mm \ Hg$,the lowering of vapour pressure will be: .............. $mm \ Hg$ (molar mass of urea $= 60 \ g \ mol^{-1}$)

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Vapour pressure in $mm \ Hg$ of $0.1 \ mole$ of urea in $180 \ g$ of water at $25^{\circ} C$ is (The vapour pressure of water at $25^{\circ} C$ is $24 \ mm \ Hg$ )

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