Calculate the relative lowering of vapour pressure of a solution containing $0.56 \ g$ of a non-volatile solute in $100 \ g$ of water [molar mass of solute $= 60 \ g \ mol^{-1}$].

The mass of a non-volatile solute of molar mass $40 \ g \ mol^{-1}$ that should be dissolved in $114 \ g$ of octane to lower its vapour pressure by $20 \%$ is (in $g$)

When a solid solute is dissolved in a volatile solvent to form a solution,then .......

What is the mole ratio of benzene $(P_B^o = 150\, torr)$ and toluene $(P_T^o = 50\, torr)$ in the vapour phase if the given solution has a vapour pressure of $120\, torr$?

"The relative lowering of the vapour pressure is equal to the mole fraction of the solute." This law is called

$X$ 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: