Dry air is passed through a solution containing $10 \ g$ of solute in $90 \ g$ of water and then through pure water. The loss in weight of the solution is $2.5 \ g$ and the loss in weight of the solvent is $0.05 \ g$. The molecular weight of the solute is:

On adding a solute to a solvent having a vapour pressure of $0.80 \ atm$,the vapour pressure reduces to $0.60 \ atm$. The mole fraction of the solute is:

In a mixture of liquids $A$ and $B$, if the mole fractions of component $A$ in vapour phase and liquid mixture are $x_1$ and $x_2$ respectively, then the total vapour pressure of liquid mixture is (where $P_{A}^{\circ}$ and $P_{B}^{\circ}$ are the vapour pressures of pure $A$ and $B$)

At $300 \ K$,the vapour pressure of a solution containing $1 \ mole$ of $n-hexane$ and $3 \ moles$ of $n-heptane$ is $550 \ mm \ Hg$. At the same temperature,if one more mole of $n-heptane$ is added to this solution,the vapour pressure of the solution increases by $10 \ mm \ Hg$. What is the vapour pressure in $mm \ Hg$ of $n-heptane$ in its pure state?

At a certain temperature,$5 \ g$ of a non-electrolyte solute is dissolved in $100 \ g$ of water. The vapour pressure of the resulting solution is $2985 \ N/m^2$,and the vapour pressure of pure water is $3000 \ N/m^2$. Calculate the molar mass of the solute.

At $25\,^{\circ}C$,the vapour pressure of pure liquid $A$ $(mol. wt. = 40)$ is $100\, torr$,while that of pure liquid $B$ is $40\, torr$ $(mol. wt. = 80)$. The vapour pressure at $25\,^{\circ}C$ of a solution containing $20\, g$ of each $A$ and $B$ is .......... $torr$.