$34.2 \ g$ of cane sugar is dissolved in $180 \ g$ of water. The relative lowering of vapour pressure will be

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

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 $= 78 \ g \ mol^{-1}$)

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 vapor pressure of an aqueous solution at $373 \, K$ is $0.925 \, atm$. What is the mole fraction of the solute?

At $300 \ K$,when a solute is added to a solvent,its vapour pressure over the mercury reduces from $50 \ mm$ to $45 \ mm$. The value of the mole fraction of the solute will be: