Vapour pressure of chloroform $(CHCl_3)$ and dichloromethane $(CH_2Cl_2)$ at $25^\circ C$ are $200 \ mm \ Hg$ and $41.5 \ mm \ Hg$ respectively. Vapour pressure of the solution obtained by mixing $25.5 \ g$ of $CHCl_3$ and $40 \ g$ of $CH_2Cl_2$ at the same temperature will be (Molecular mass of $CHCl_3 = 119.5 \ u$ and molecular mass of $CH_2Cl_2 = 85 \ u$)

  • A
    $173.9 \ mm \ Hg$
  • B
    $615.0 \ mm \ Hg$
  • C
    $347.9 \ mm \ Hg$
  • D
    None of the above

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The vapour pressure of a solvent decreases by $2.5 \ mm \ Hg$ by adding a solute. What is the mole fraction of solute? (Vapour pressure of pure solvent is $250 \ mm \ Hg$)

Assertion $(A)$: The vapour pressure of $0.1 \ M$ sugar solution is less than that of $0.1 \ M$ $KCl$ solution.
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The correct answer 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 pressures of $A$ and $B$ at $25^{\circ} C$ are $90 \ mm \ Hg$ and $15 \ mm \ Hg$ respectively. If $A$ and $B$ are mixed such that the mole fraction of $A$ in the mixture is $0.6$,then the mole fraction of $B$ in the vapour phase is $x \times 10^{-1}$. The value of $x$ is $.....$ (Nearest integer)

Relative lowering of vapour pressure of a dilute solution of glucose dissolved in $1 \ kg$ of water is $0.002$. The molality of the solution is (in $m$)

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