The vapour pressure of pure $CHCl_3$ and $CH_2Cl_2$ are $200 \,atm$ and $41.5 \,atm$ respectively. The weights of $CHCl_3$ and $CH_2Cl_2$ are respectively $11.9 \,g$ and $17 \,g$. The vapour pressure of the solution will be (in $,atm$)

The weight in grams of a non-volatile solute (mol. wt. $60$) to be dissolved in $90 \ g$ of water to produce a relative lowering of vapour pressure of $0.02$ is

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

Vapour pressure in $mm \ Hg$ of $0.1 \ mole$ of urea in $180 \ g$ of water at $25^{\circ} C$ is (The vapour pressure of water at $25^{\circ} C$ is $24 \ mm \ Hg$ )

The boiling points of $C_6H_6, CH_3OH, C_6H_5NH_2$ and $C_6H_5NO_2$ are $80 \ ^oC, 65 \ ^oC, 184 \ ^oC$ and $212 \ ^oC$ respectively. Which will show the highest vapour pressure at room temperature?

Calculate the vapour pressure of the solution if the relative lowering of vapour pressure and the vapour pressure of the pure solvent are $0.018$ and $18 \ mm \ Hg$ respectively at $300 \ K$. (in $mm \ Hg$)