Dry air was passed successively through a solution of $5 \ g$ of a solute in $80 \ g$ of water and then through pure water. The loss in weight of solution was $2.50 \ g$ and that of pure solvent $0.04 \ g$. What is the molecular weight of the solute?

On adding a solute to a solvent having a vapour pressure of $0.80 \ atm$,the vapour pressure reduces to $0.60 \ atm$. The mole fraction of the solute is:

$A$ solution is obtained by dissolving $12 \ g$ of urea (mol.wt. $60$) in a litre of water. Another solution is obtained by dissolving $68.4 \ g$ of cane sugar (mol.wt. $342$) in a litre of water at the same temperature. The lowering of vapour pressure in the first solution is

What is the mass in $g$ of a non-volatile solute with a molecular weight of $40$ that should be dissolved in $57 \ g$ of octane to reduce its vapour pressure to $80 \%$ of its original value?

Assertion $:-$ Vapour pressure of a liquid increases on increasing surface area.
Reason $:-$ Rate of vaporisation of a liquid is directly proportional to surface area.

$X$ is a non-volatile solute and $Y$ is a volatile solvent. The following vapour pressures are observed by dissolving $X$ in $Y$ at different concentrations:
| $X / \text{mol L}^{-1}$ | $Y / \text{mm of Hg}$ |
| :--- | :--- |
| $0.10$ | $p_1$ |
| $0.25$ | $p_2$ |
| $0.01$ | $p_3$ |
The correct order of vapour pressures is: