"Relative lowering in vapour pressure of a solution containing a non-volatile solute is directly proportional to the mole fraction of the solute". The above statement is:

At $298 \ K$,the vapour pressure of a solution of $7.5 \ g$ of non-volatile solute in $90 \ g$ of water is $2.8 \ kPa$. If $18 \ g$ of water is added to this solution,the vapour pressure becomes $2.81 \ kPa$ at the same temperature. The molar mass of the solute in $g \ mol^{-1}$ is:

$X$ 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:

The vapour pressure of water at room temperature is lowered by $5\%$ by dissolving a solute in it. Then the approximate molality of the solution is

At $300 \ K$,when a solute is added to a solvent,its vapour pressure over the mercury reduces from $50 \ mm$ to $45 \ mm$. The value of the mole fraction of the solute will be:

The vapour pressure of pure benzene at a certain temperature is $0.850 \ bar$. $A$ non-volatile,non-electrolyte solid weighing $0.5 \ g$ when added to $39.0 \ g$ of benzene (molar mass $78 \ g \ mol^{-1}$),the vapour pressure of the solution becomes $0.845 \ bar$. What is the molar mass of the solid substance?