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 :

$A$ solution is made by mixing $1 \ mole$ of volatile liquid $A$ with $3 \ moles$ of volatile liquid $B$. The vapour pressure of pure $A$ is $200 \ mm \ Hg$ and that of the solution is $500 \ mm \ Hg$. The vapour pressure of pure $B$ and the least volatile component of the solution,respectively,are $:$

$224 \, mL$ of $SO_{2(g)}$ at $273 \, K$ and $1 \, atm$ is passed through $100 \, mL$ of $0.1 \, M \, NaOH$ solution. The non-volatile solute produced is dissolved in $36 \, g$ of water. The lowering of vapour pressure of the solution (assuming the solution is dilute) $(P^0_{H_2O} = 24 \, mm \, Hg)$ is $x \times 10^{-2} \, mm \, Hg$. The value of $x$ is ...... . (Integer answer)

At $T \ K$, $0.1 \ mol$ of a non-volatile solute was dissolved in $0.9 \ mol$ of a volatile solvent. The vapour pressure of pure solvent is $0.9 \ bar$. What is the vapour pressure (in $bar$) of the solution?

The vapor pressures of two liquids $A$ and $B$ are $80 \ mm$ and $60 \ mm$ respectively. If $3 \ mol$ of $A$ and $2 \ mol$ of $B$ are mixed,the total vapor pressure of the resulting solution will be .......... $mm$.

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