How many grams of a non-volatile solute with a molar mass of $90$ must be dissolved in $97.5 \ g$ of water to cause a $2.5 \%$ decrease in vapour pressure?

The vapour pressure of an aqueous solution of methanol is:

At $25\,^oC$,the vapour pressure of $CCl_4$ is $143\,mm\,Hg$. If $0.5\,g$ of a non-volatile solute (molar mass $= 65\,g/mol$) is dissolved in $100\,mL$ of $CCl_4$,what will be the vapour pressure of the solution? (Density of $CCl_4 = 1.58\,g/cm^3$)

The vapour pressure of pure benzene and toluene are $160$ $mm \ Hg$ and $60$ $mm \ Hg$ respectively. The mole fraction of benzene in the vapour phase in contact with an equimolar solution of benzene and toluene is:

$2\, \text{moles}$ of liquid $A$ is mixed with $3\, \text{moles}$ of liquid $B$. Calculate the mole fraction of $A$ in the vapour phase if $P_A^o = 100\, \text{mm}$ and $P_B^o = 200\, \text{mm}$.

$18 \ g$ of glucose is dissolved in $90 \ g$ of water. The relative lowering of vapour pressure of the solution is equal to