Assertion $:-$ Vapour pressure of a liquid increases on increasing surface area.
Reason $:-$ Rate of vaporisation of a liquid is directly proportional to surface area.

The vapour pressure of benzene at a certain temperature is $640 \ mm$ of $Hg$. $A$ non-volatile and non-electrolyte solid weighing $2.175 \ g$ is added to $39.08 \ g$ of benzene. If the vapour pressure of the solution is $600 \ mm$ of $Hg$,what is the molecular weight of the solid substance?

Two $5 \ molal$ solutions are prepared by dissolving a non-electrolyte,non-volatile solute separately in the solvents $X$ and $Y$. The molecular weights of the solvents are $M_X$ and $M_Y$,respectively,where $M_X = \frac{3}{4} M_Y$. The relative lowering of vapour pressure of the solution in $X$ is $m$ times that of the solution in $Y$. Given that the number of moles of solute is very small in comparison to that of solvent,the value of $m$ is

If the vapor pressure of liquid $A$ is ${P_A}^o$ and the vapor pressure of liquid $B$ is ${P_B}^o$,then the partial vapor pressure of $A$ in the solution will be .....

$3 \ g$ of urea is dissolved in $45 \ g$ of $H_2O$. The relative lowering in vapour pressure is

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.]