An aqueous solution of $2\%$ non-volatile solute exerts a pressure of $1.004 \, bar$ at the normal boiling point of the solvent. The molar mass of the solute is $......... \, g/mol$.

The vapour pressures of two volatile liquids $A$ and $B$ at $25^{\circ} C$ are $50 \ Torr$ and $100 \ Torr,$ respectively. If the liquid mixture contains $0.3$ mole fraction of $A$,then the mole fraction of liquid $B$ in the vapour phase is $\frac{x}{17}$. The value of $x$ is $...$

Calculate the amount of solute dissolved in $612 \ g$ of water at $30^{\circ} C$ if molar mass of solute is $342 \ g \ mol^{-1}$ (Relative vapour pressure lowering is $0.025$ and molar mass of water $18 \ g \ mol^{-1}$ ). (in $g$)

At $T \ K$,the vapour pressure of an aqueous solution of a non-volatile solute,whose mole fraction is $0.02$,is found to be $34.65 \ mm \ Hg$. What is the vapour pressure (in $mm \ Hg$) of pure water at the same temperature?

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:

Two liquids $A$ and $B$ have vapour pressure in the ratio $P_A^o : P_B^o = 1 : 3$ at a certain temperature. If the ratio of mole fractions of $A$ to $B$ in the vapour phase is $4 : 3$,then the mole fraction of $B$ in the solution at the same temperature is