$6 \ g$ of a non-volatile solute $(x)$ is dissolved in $100 \ g$ of water. The relative lowering of vapour pressure of the resultant solution is $0.006$. What is the molar mass (in $g \ mol^{-1}$) of $x$?

An aqueous solution containing $28\%$ by mass of a liquid $A$ (molar mass $= 140 \ g/mol$) has a vapour pressure of $160 \ mm$ at $37 \ ^\circ C$. Find the vapour pressure of the pure liquid $A$ approximately .......... $mm$. (The vapour pressure of pure water at $37 \ ^\circ C$ is $150 \ mm$)

"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:

The mass of non-volatile,non-electrolyte solute (molar mass $= 50 \ g \ mol^{-1}$) needed to be dissolved in $114 \ g$ octane to reduce its vapour pressure to $75\%$ is .............. $g$.

At $T \ K$, $0.1 \ mol$ of a non-volatile solute was dissolved in $0.9 \ mol$ of a volatile solvent. The vapour pressure of pure solvent is $0.9 \ bar$. What is the vapour pressure (in $bar$) of the solution?