$9 \ g$ anhydrous oxalic acid (mol. $Wt. = 90$) was dissolved in $9.9 \ moles$ of water. If vapour pressure of pure water is $P_1^o$,the vapour pressure of solution is (in $P_1^o$)

Relative lowering of vapour pressure of a dilute solution is $0.5$. What is the mole fraction of the non-volatile solute?

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:

The vapour pressure of water at room temperature is lowered by $5\%$ by dissolving a solute in it. Then the approximate molality of the solution is

How many grams of a non-volatile solute (molar mass $40 \ g/mol$) must be added to $114 \ g$ of octane to reduce its vapour pressure by $20\%$ (in $g$)?

At $300 \ K$, $0.06 \ kg$ of an organic solute is dissolved in $1 \ kg$ water. The vapour pressure of the solution at $300 \ K$ is $3.768 \ kPa$. If the vapour pressure of water at that temperature is $3.78 \ kPa$, what is the molar mass of the organic solute (in $g \ mol^{-1}$)? (Assume the solution is dilute)