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 $mm \ Hg$)

What weight of glucose must be dissolved in $100 \ g$ of water to lower the vapour pressure by $0.20 \ mm \ Hg$ (in $g$)? (Assume dilute solution is being formed) Given: Vapour pressure of pure water is $54.2 \ mm \ Hg$ at room temperature. Molar mass of glucose is $180 \ g \ mol^{-1}$

The vapour pressure of a pure solvent is $0.70 \ atm.$ When a non-volatile solute $A$ is dissolved in it,the vapour pressure of the solution becomes $0.35 \ atm.$ The mole fraction of solute $A$ will be$:-$

The vapour pressure of water at $23^{\circ} C$ is $19.8 \ mm$. If $0.1 \ mole$ of glucose is dissolved in $178.2 \ g$ of water, what is the vapour pressure (in $mm$) of the resultant solution?

The vapour pressures of pure liquids $A$ and $B$ are $400$ and $600\, mm\, Hg$,respectively at $298\, K$. On mixing the two liquids,the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid $B$ is $0.5$ in the mixture. The vapour pressure of the final solution,the mole fractions of components $A$ and $B$ in vapour phase,respectively are

An aqueous solution of $2 \%$ non-volatile solute exerts a pressure of $1.004 \ bar$ at the normal boiling point of the solvent. What is the molar mass of the solute?