$A$ liquid is kept in a closed vessel. If a glass plate (negligible mass) with a small hole is kept on top of the liquid surface,then the vapour pressure of the liquid in the vessel is
- AMore than what would be if the glass plate were removed
- BSame as what would be if the glass plate were removed
- CLess than what would be if the glass plate were removed
- DCannot be predicted
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How many grams of a non-volatile solute with a molar mass of $90$ must be dissolved in $97.5 \ g$ of water to cause a $2.5 \%$ decrease in vapour pressure?
Medium
View SolutionAt $T \ K$,the vapour pressure of an $x$ molal aqueous solution containing a non-volatile solute is $12.078 \ kPa$. The vapour pressure of pure water at $T \ K$ is $12.3 \ kPa$. What is the value of $x$?
"Relative lowering in vapour pressure of a solution containing a non-volatile solute is directly proportional to the mole fraction of the solute". The above statement is:
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View SolutionWhen a non-volatile solute is dissolved in a solvent,the relative lowering of vapour pressure is equal to
Vapour pressure of chloroform $(CHCl_{3})$ and dichloromethane $(CH_{2}Cl_{2})$ at $298\,K$ are $200\,mm\,Hg$ and $415\,mm\,Hg$ respectively. $(i)$ Calculate the vapour pressure of the solution prepared by mixing $25.5\,g$ of $CHCl_{3}$ and $40\,g$ of $CH_{2}Cl_{2}$ at $298\,K$ and,$(ii)$ mole fractions of each component in vapour phase.
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