Relative lowering in vapour pressure of a solution containing a non-volatile solute is the ratio of:
- ANumber of moles of solute to number of moles of solvent.
- BNumber of moles of solvent to total number of moles of solution.
- CNumber of moles of solvent to number of moles of solute.
- DNumber of moles of solute to total number of moles of solution.
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Similar Questions
How many grams of glucose are required to prepare an aqueous solution of glucose having a vapor pressure of $23.324 \ mm \ Hg$ at $25^{\circ} C$ in $100 \ g$ of water? The vapor pressure of pure water at $25^{\circ} C$ is $23.8 \ mm \ Hg$.
The vapour pressure of water at $20\,^{\circ}C$ is $17.54\,mm$. When $20\,g$ of a non-ionic substance is dissolved in $100\,g$ of water,the vapour pressure is lowered by $0.30\,mm$. What is the molecular weight of the substance?
Medium
View SolutionIf $p^{\circ}$ and $p$ are the vapour pressure of the pure solvent and solution, and $n_{1}$ and $n_{2}$ are the moles of solute and solvent respectively in the solution, then the correct relation between $p$ and $p^{\circ}$ is:
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View SolutionVapour pressure of a solution containing $18 \ g$ of glucose and $178.2 \ g$ of water at $100^{\circ} C$ is (Vapour pressure of pure water at $100^{\circ} C = 760 \ torr$) (in $torr$)
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View Solution$A$ solution is prepared by dissolving $68 \text{ g}$ of sucrose $(C_{12}H_{22}O_{11})$ in $1 \text{ kg}$ of water. Calculate the vapour pressure of the solution at $298 \text{ K}$. Given: Vapour pressure of pure water at $298 \text{ K} = 18 \text{ mm Hg}$ and mole fraction of solvent $(x_1)$ $= 0.9964$. (in $\text{ mm Hg}$)
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