One mole of glucose is dissolved in $2 \, moles$ of water. The vapour pressure of the solution relative to that of water is

What is the vapour pressure of a solution containing $1 \ mol$ of a non-volatile solute in $36 \ g$ of water (in $mm \ Hg$)? $(P_1^0 = 400 \ mm \ Hg)$

Calculate the vapour pressure of pure volatile liquid $A$ at a given temperature if the mole fraction and vapour pressure of pure volatile liquid $B$ are $0.4$ and $900 \ mm \ Hg$ respectively,given that the total vapour pressure of the solution is $600 \ mm \ Hg$. (in $mm \ Hg$)

The ratio of loss in weight of solvent to gain in weight of $CaCl_{2}$ tube in Ostwald-Walker method is:

At $300 \ K$,the vapour pressures of two pure liquids,$A$ and $B$ are $100 \ mm \ Hg$ and $500 \ mm \ Hg$,respectively. If in a mixture of $A$ and $B$,the total vapour pressure is $300 \ mm \ Hg$,the mole fractions of $A$ in the liquid and in the vapour phase,respectively,are:

Two beakers $A$ and $B$ containing pure water and an aqueous solution,respectively,are placed in a closed container. What will be observed gradually with the passage of time?