The vapour pressure of acetone at $20\,^{\circ}C$ is $185\,torr.$ When $1.2\,g$ of a non-volatile substance was dissolved in $100\,g$ of acetone at $20\,^{\circ}C,$ its vapour pressure was $183\,torr.$ The molar mass $(g\,mol^{-1})$ of the substance is:

Vapour pressure of chloroform $(CHCl_3)$ and dichloromethane $(CH_2Cl_2)$ at $25^\circ C$ are $200 \ mm \ Hg$ and $41.5 \ mm \ Hg$ respectively. Vapour pressure of the solution obtained by mixing $25.5 \ g$ of $CHCl_3$ and $40 \ g$ of $CH_2Cl_2$ at the same temperature will be (Molecular mass of $CHCl_3 = 119.5 \ u$ and molecular mass of $CH_2Cl_2 = 85 \ u$)

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 :

Relative lowering of vapour pressure of a dilute solution of glucose dissolved in $1 \ kg$ of water is $0.002$. The molality of the solution is (in $m$)

At $80\,^{\circ}C$,the vapour pressure of pure liquid '$A$' is $520\, mm\, Hg$ and that of pure liquid '$B$' is $1000\, mm\, Hg$. If a mixture solution of '$A$' and '$B$' boils at $80\,^{\circ}C$ and $1\, atm$ pressure,then the amount of '$A$' in the mixture is ......... $mol\, \%$. $(1\, atm = 760\, mm\, Hg)$

When a non-volatile solute is added to the solvent,the vapour pressure of the solvent decreases by $10 \ mm \ Hg$. The mole fraction of the solute in the solution is $0.2$. What would be the mole fraction of the solvent if the decrease in vapour pressure is $20 \ mm \ Hg$?