The correct option for the value of vapour pressure of a solution at $45^{\circ} C$ with benzene to octane in molar ratio $3 : 2$ is ...... $mm$ of $Hg$. [At $45^{\circ} C$,vapour pressure of benzene is $280 \ mm \ Hg$ and that of octane is $420 \ mm \ Hg$. Assume ideal solution.]

  • A
    $160$
  • B
    $168$
  • C
    $336$
  • D
    $350$

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When $25 \ g$ of a non-volatile solute is dissolved in $100 \ g$ of water, the vapour pressure is lowered by $2.25 \times 10^{-1} \ mm$. If the vapour pressure of water at $20^{\circ}C$ is $17.5 \ mm$, what is the molecular weight of the solute?

The vapour pressure of a pure liquid $A$ is $70 \ torr$ at $300 \ K$. It forms an ideal solution with another liquid $B$. The mole fraction of $B$ is $0.2$ and the total vapour pressure of the solution is $84 \ torr$ at the same temperature. The vapour pressure of pure liquid $B$ (in $torr$) 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:

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$A$ liquid is kept in a closed vessel. If a glass plate (negligible mass) with a small hole is kept on top of the liquid surface,then the vapour pressure of the liquid in the vessel is

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