Question

46. The weight of urea to be dissolved in 100 g.
of water to decrease the vapour pressure of
water by 5% is
1) 20 g.
2) 14.66 g
3) 15.24 g
4)16.66 g.​

Answer 1

Given,

• $\bold{w_1=100\:g}$

• $\bold{w_2=?}$

• $\bold{m_1=18\:g}$

• $\bold{m_2=60\:g}$

• ∆P/P° = 5%

Lowering of vapor pressure by non-volatile solute is formulated by ,

$\boxed{\bold{\frac{\Delta P}{P}=\frac{w_2m_1}{w_1m_2}}}$

$\bold{\frac{5}{100}=\frac{w_2\times 18}{100\times 6}}$

$\bold{w_2=\frac{5\times 100\times 6}{100\times 18}}$

$\bold{w_2=1.67\:g}$

Therefore,

1.67 g of urea is required to lower the vapour pressure.