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

$x \ g$ of a solute is dissolved in $y \ g$ of two different liquids $A$ and $B$. The relative lowering of vapor pressure of the solution in $A$ is twice that of the solution in $B$. If the moles of solute are negligible compared to the moles of solvent,which of the following statements regarding the molar masses $M_A$ and $M_B$ of the solvents is correct?

The vapour pressure of $30 \%$ $(w/v)$ aqueous solution of glucose is $...... \ mm \ Hg$ at $25^{\circ} \ C$. [Given : The density of $30 \%$ $(w/v)$ aqueous solution of glucose is $1.2 \ g \ cm^{-3}$ and vapour pressure of pure water is $24 \ mm \ Hg$.] (Molar mass of glucose is $180 \ g \ mol^{-1}$.)

$X$ 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 pure volatile liquid $A$ at a given temperature if the mole fraction and vapour pressure of pure volatile liquid $B$ are $0.4$ and $900 \ mm \ Hg$ respectively,given that the total vapour pressure of the solution is $600 \ mm \ Hg$. (in $mm \ Hg$)

$18 \ g$ of glucose is dissolved in $90 \ g$ of water. The relative lowering of vapour pressure of the solution is equal to