If $P_0$ and $P_s$ are the vapour pressures of the solvent and its solution respectively,and $N_1$ and $N_2$ are the mole fractions of the solvent and solute respectively,then which of the following is correct?

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

If $x \ g$ of a non-volatile,non-electrolytic solute is dissolved in $114 \ g$ of octane to reduce its vapour pressure to $80 \%$ of that of pure octane,then $x$ is ........... $g$. (Molar mass of solute $= 40 \ g/mol$)

The vapour pressure of pure liquids $A$ and $B$ are $450$ $mm$ $Hg$ and $700$ $mm$ $Hg$ respectively,at $350$ $K$. Find out the composition of the liquid mixture if the total vapour pressure is $600$ $mm$ $Hg$. Also,find the composition of the vapour phase.

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 :

Dry air was passed successively through a solution of $5 \ g$ of a solute in $80 \ g$ of water and then through pure water. The loss in weight of solution was $2.50 \ g$ and that of pure solvent $0.04 \ g$. What is the molecular weight of the solute?