Calculate the vapour pressure of an aqueous solution which contains 5 mass per cent of urea. The vapour pressure of pure water is 23.5 mm Hg. Molar mass of urea is 60.

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Solution

Mass of urea = 5 g
Mass of water = (100-5) = 95 g
No. of moles of urea $= \frac{5}{60} = 0.083$
No. of moles of water $= \frac{95}{18} = 5.278$
Total number of moles $= 5.278 + 0.083$
$= 5.361$
Mole fraction of solvent $= \frac{5.278}{5.361}$
$P_{s} =$ Mole fraction of solvent $\times P_{0}$
$= \frac{5.278}{5.361} \times 23.5 = 23.14 m m$

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Calculate the vapour pressure of an aqueous solution which contains 5 mass per cent of urea. The vapour pressure of pure water is 23.5 mm Hg. Molar mass of urea is 60.

Mass of urea = 5 gMass of water = (100-5) = 95 gNo. of moles of urea =560=0.083No. of moles of water =9518=5.278Total number of moles =5.278+0.083 =5.361Mole fraction of solvent =5.2785.361Ps= Mole fraction of solvent ×P0=5.2785.361×23.5=23.14mm