$A$ solution contains $KCl$ in water. It is heated continuously and the vapour formed is removed continuously. The vapour pressure of the solution will be:

$A$ set of solutions is prepared using $180 \ g$ of water as a solvent and $10 \ g$ of different non-volatile solutes $A, B$ and $C$. The relative lowering of vapour pressure in the presence of these solutes are in the order
[Given,molar mass of $A = 100 \ g \ mol^{-1}; B = 200 \ g \ mol^{-1}; C = 10,000 \ g \ mol^{-1}]$

$X$ 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:

Calculate the vapour pressure of a solution containing a mixture of $2 \ mol$ of volatile liquid $A$ and $3 \ mol$ of volatile liquid $B$ at room temperature. $(P_{A}^{\circ} = 420 \ mm \ Hg, P_{B}^{\circ} = 610 \ mm \ Hg)$ (in $mm \ Hg$)

The partial vapour pressure of any volatile component of a solution is equal to the vapour pressure of the pure component multiplied by its mole fraction in the solution is called:

At $300 \ K$,the vapour pressures of two pure liquids,$A$ and $B$ are $100 \ mm \ Hg$ and $500 \ mm \ Hg$,respectively. If in a mixture of $A$ and $B$,the total vapour pressure is $300 \ mm \ Hg$,the mole fractions of $A$ in the liquid and in the vapour phase,respectively,are: