What is the vapour pressure of a solution containing $0.1 \ mol$ of non-volatile solute dissolved in $16.2 \ g$ of water (in $mm \ Hg$)? $(P_1^{\circ} = 32 \ mm \ Hg)$

The vapour pressure of a given liquid decreases if?

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 .............. $g$.

$X$ 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 pure $CCl_4$ (molar mass $= 154 \ g \ mol^{-1}$) and $SnCl_4$ (molar mass $= 170 \ g \ mol^{-1}$) at $25^{\circ} C$ are $115.0 \ torr$ and $238.0 \ torr$ respectively. Assuming ideal behaviour,calculate the total approximate vapour pressure in $torr$ of a solution containing $10 \ g$ of $CCl_4$ and $15 \ g$ of $SnCl_4$.

$A$ solution is prepared by mixing $8.5 \ g$ of $CH_2Cl_2$ and $11.95 \ g$ of $CHCl_3$. If the vapour pressures of pure $CH_2Cl_2$ and $CHCl_3$ at $298 \ K$ are $415 \ mm \ Hg$ and $200 \ mm \ Hg$ respectively,the mole fraction of $CHCl_3$ in the vapour phase is: (Molar mass of $Cl = 35.5 \ g \ mol^{-1}$)