For a dilute solution,Raoult's law states that

  • A
    The lowering of vapour pressure is equal to mole fraction of solute
  • B
    The relative lowering of vapour pressure is equal to mole fraction of solute
  • C
    The relative lowering of vapour pressure is proportional to the amount of solute in solution
  • D
    The vapour pressure of the solution is equal to the mole fraction of solvent

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$160 \ g$ of non-volatile solute '$A$' is dissolved in $54 \ mL$ of water at $373 \ K$. What is the vapour pressure of the aqueous solution of '$A$' (in $Torr$)? (Given: molecular weight of '$A$' = $160 \ g \ mol^{-1}$)

$A$ solution is prepared by dissolving $68 \text{ g}$ of sucrose $(C_{12}H_{22}O_{11})$ in $1 \text{ kg}$ of water. Calculate the vapour pressure of the solution at $298 \text{ K}$. Given: Vapour pressure of pure water at $298 \text{ K} = 18 \text{ mm Hg}$ and mole fraction of solvent $(x_1)$ $= 0.9964$. (in $\text{ mm Hg}$)

At a given temperature,the vapour pressure of a solution of two volatile liquids $A$ and $B$ is given by the equation $P_S = 150 - 60 X_B$ (where $X_B$ is the mole fraction of $B$). The vapour pressures of pure $A$ and pure $B$ at the same temperature are respectively:

Vapour pressure of chloroform $(CHCl_{3})$ and dichloromethane $(CH_{2}Cl_{2})$ at $298\,K$ are $200\,mm\,Hg$ and $415\,mm\,Hg$ respectively. $(i)$ Calculate the vapour pressure of the solution prepared by mixing $25.5\,g$ of $CHCl_{3}$ and $40\,g$ of $CH_{2}Cl_{2}$ at $298\,K$ and,$(ii)$ mole fractions of each component in vapour phase.

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Calculate the relative lowering of vapour pressure if the vapour pressure of benzene and vapour pressure of solution of non-volatile solute in benzene are $640 \ mmHg$ and $590 \ mmHg$ respectively at the same temperature.

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