Vapour pressure of water is $360$ mm Hg, how much urea should be added to $200$ mL water to reduce its vapour pressure by $0.5 %$ ? (Molecular wt. of urea $= 60$ )

2.52 g

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3.33 g

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3.96 g

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None of these

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Solution

The correct option is A $3.33 g$
Vapour pressure of pure water = $p^{o}$ = 360 mm Hg
Lowered vapour pressure = p = 360 - 0.5 % of 360 = 360 - 1.8 = 358.2 mm Hg
Now,
weight of water = w1 = $200 m l \times 1 g / m l$ = 200 g
Weight of urea = w2 = ?
Molecular weight of water M1 = 18 g/mol
Molecular weight of urea M2 = 60 g/mol
According to Raoult's law:
$\frac{p^{o} - p}{p} = \frac{n_{2}}{n_{1} + n_{2}}$

$\frac{360 - 358.2}{358.2} = \frac{\frac{w_{2}}{M_{2}}}{\frac{w_{1}}{M_{1}} + \frac{w_{2}}{M_{2}}}$

$0.005 = \frac{\frac{w_{2}}{M_{2}}}{\frac{w_{1}}{M_{1}} + \frac{w_{2}}{M_{2}}}$

$0.005 [\frac{w_{1}}{M_{1}} + \frac{w_{2}}{M_{2}}] = \frac{w_{2}}{M_{2}}$

$0.005 \times \frac{w_{1}}{M_{1}} = \frac{w_{2}}{M_{2}} - 0.005 \times \frac{w_{2}}{M_{2}}$

$0.005 \times \frac{w_{1}}{M_{1}} = 0.995 \times \frac{w_{2}}{M_{2}}$

$\frac{0.005}{0.995} \times \frac{w_{1}}{M_{1}} \times M_{2} = w_{2}$

$w_{2} = 0.005 \times \frac{200}{18} \times 60$

$w_{2} = 3.33 g$
Hence, 3.33 g of urea must be added.

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