The vapour pressure of two pure liquids $(A)$ and $(B)$ are $100 \ torr$ and $80 \ torr$ respectively. The total pressure of the solution obtained by mixing $2 \ mole$ of $(A)$ and $3 \ mole$ of $(B)$ would be ........ $torr$.

The vapour pressure of a pure liquid $A$ is $70 \ torr$ at $300 \ K$. It forms an ideal solution with another liquid $B$. The mole fraction of $B$ is $0.2$ and the total vapour pressure of the solution is $84 \ torr$ at the same temperature. The vapour pressure of pure liquid $B$ (in $torr$) is:

The mass of glucose that should be dissolved in $100 \ g$ of water in order to produce the same relative lowering of vapour pressure as is produced by dissolving $1 \ g$ of urea (molar mass $= 60 \ g/mol$) in $50 \ g$ of water is ........... $g$. (Assume dilute solution in both cases):-

When $1 \ g$ of urea is dissolved in $50 \ g$ of water,the relative lowering of vapour pressure of the solution is equal to the relative lowering of vapour pressure of a solution formed by dissolving glucose in the same amount of water. How many grams of glucose were dissolved?

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

If $P^o$ and $P$ are the vapour pressure of a solvent and its solution respectively and $N_1$ and $N_2$ are the mole fractions of the solvent and solute respectively,then the correct relation is: