$A$ solution containing $30 \, g$ of non-volatile solute in exactly $90 \, g$ water has a vapour pressure of $21.85 \, mm \, Hg$ at $25 \, ^oC$. Further $18 \, g$ of water is then added to the solution. The resulting solution has a vapour pressure of $22.15 \, mm \, Hg$ at $25 \, ^oC$. Calculate the molecular weight of the solute.

An open beaker of water in equilibrium with water vapour is in a sealed container. When a few grams of glucose are added to the beaker of water,the rate at which water molecules :

Two liquids $A$ and $B$ form an ideal solution. At $320 \ K$, the vapour pressure of the solution, containing $3 \ mol$ of $A$ and $1 \ mol$ of $B$ is $500 \ mm \ Hg$. At the same temperature, if $1 \ mol$ of $A$ is further added to this solution, the vapour pressure of the solution increases by $20 \ mm \ Hg$. The vapour pressure (in $mm \ Hg$) of $B$ in the pure state is . . . . . . (Nearest integer).

At $50^{\circ} C$, the vapour pressure of pure benzene is $268 \ torr$. The number of moles of non-volatile solute per mole of benzene required to prepare a solution having a vapour pressure of $167 \ torr$ at the same temperature is (molar mass of benzene $= 78 \ g \ mol^{-1}$)

Vapour pressure of solvent is $17.5 \ mm \ Hg$ while that of dilute solution is $17.45 \ mm \ Hg$; the mole fraction of solvent is

The vapour pressure of pure benzene and methyl benzene at $27^{\circ} C$ is given as $80 \ Torr$ and $24 \ Torr$,respectively. The mole fraction of methyl benzene in vapour phase,in equilibrium with an equimolar mixture of those two liquids (ideal solution) at the same temperature is. . . . . . .$\times 10^{-2}$ (nearest integer)