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Home » Find the vapor pressure of water and its relative lowering in the solution which is 50 g of urea (NH2CONH2) dissolved in 850 g of water. (Vapor pressure of pure water at 298 K is 23.8 mm Hg)
Find the vapor pressure of water and its relative lowering in the solution which is 50 g of urea (NH2CONH2) dissolved in 850 g of water. (Vapor pressure of pure water at 298 K is 23.8 mm Hg)
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Find the vapor pressure of water and its relative lowering in the solution which is 50 g of urea (NH2CONH2) dissolved in 850 g of water. (Vapor pressure of pure water at 298 K is 23.8 mm Hg)

As we know that vapour pressure of water, P_{1}^{\infty}=23.8 \mathrm{~mm} of \mathrm{Hg}

Weight of water, w_{1}=850 \mathrm{~g}

Weight of urea, w_{2}=50 \mathrm{~g}

Molecular weight of water, M_{1}=18 \mathrm{~g} \mathrm{~mol}^{-1}

Molecular weight of urea, M_{2}=60 \mathrm{~g} \mathrm{~mol}^{-1}

We must now determine the vapour pressure of water in the solution. Vapour pressure is denoted by.

As a result of Raoult’s law, we now have:

\frac{P_{1}^{\circ}-P_{1}}{P_{1}^{\circ}}=\frac{n_{2}}{n_{1}+n_{2}}

\Rightarrow \frac{23.8-{{P}_{1}}}{23.8}=\frac{0.83}{47.22+0.83}

\Rightarrow \frac{23.8-{{P}_{1}}}{23.8}=0.0173

\Rightarrow P_{1}=23.4 \mathrm{~mm} of \mathrm{Hg} As a result, the vapour pressure of water in the given solution is 23.4 \mathrm{~mm} of \mathrm{Hg} and its relative lowering is 0.0173.

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