What is the molar mass of a solute when $2.3 \ g$ of a non-volatile solute is dissolved in $46 \ g$ of benzene at $30^{\circ} C$? (Relative lowering of vapour pressure is $0.06$ and the molar mass of benzene is $78 \ g \ mol^{-1}$)

  • A
    $65 \ g \ mol^{-1}$
  • B
    $80 \ g \ mol^{-1}$
  • C
    $72 \ g \ mol^{-1}$
  • D
    $48 \ g \ mol^{-1}$

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At a given temperature,the vapour pressure of a solution of two volatile liquids $A$ and $B$ is given by the equation $P_S = 120 - 80 \, X_B$ (where $X_B$ is the mole fraction of $B$). The vapour pressures of pure $A$ and $B$ at the same temperature are respectively:

$A$ 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 $Cl = 35.5 \ g \ mol^{-1}$)

Vapour pressure of a solution and of a pure solvent are $P_1$ and $P_1^0$ respectively. If $\frac{P_1}{P_1^0}$ is $0.15$,find the mole fraction of the solute.

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