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 weight of urea=60)

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Solution

Vapour pressure of pure water = Po = 360 mm/g
Lowered Vapour pressure P = 360-0.5 of 360
= 360.18
= 358.2 mm Hg
weight of water $= 200 m l \times 1 g m = 200 g$ .
weight of water w2 = ?
Molecular weight of water (m_{1}) = 18 g/mol
Molecular weight of water (m_{2}) = 60g/mol
According to Rout's law
$\frac{P O - P}{P} = \frac{n 2}{n 1 + n 2}$
$360 - 358.2 = w_{2} / m_{2}$ divided by $w_{1} / m_{1} + w_{2} / m$
$0.005 = \frac{w_{2} / m_{2}}{w_{1} / m_{1} + w_{2} / m_{2}}$
$\frac{0.005}{0.995} \times w_{1} / m_{1} \times m_{2} = w_{2}$
$\frac{0.005 \times 200}{18 \times 60} = w_{2}$
$3.33 g = w_{2}$
3.33g area should be added

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