(B) According to Raoult's law,the relative lowering of vapour pressure is equal to the mole fraction of the solute: $\frac{\Delta p}{p^o} = x_{solute}

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(B) According to Raoult's law,the relative lowering of vapour pressure is equal to the mole fraction of the solute: $\frac{\Delta p}{p^o} = x_{solute}$.Given $\Delta p = 10 \ mm \ Hg$ and $x_{solute} = 0.2$,we have $\frac{10}{p^o} = 0.2$,which gives $p^o = \frac{10}{0.2} = 50 \ mm \ Hg$.For the second case,where $\Delta p = 20 \ mm \ Hg$,the new mole fraction of the solute $(x'_{solute})$ is $\frac{20}{p^o} = \frac{20}{50} = 0.4$.Since the sum of mole fractions of solute and solvent is $1$,the mole fraction of the solvent is $x_{solvent} = 1 - x'_{solute} = 1 - 0.4 = 0.6$.

Class 12 Chemistry Solutions Lowering of vapour pressure

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The vapour pressure of a solvent decreases by $10 \ mm$ of $Hg$ when a non-volatile solute is added to the solvent. The mole fraction of the solute in the solution is $0.2$. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be $20 \ mm$ of $Hg$?

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AIPMT 1998 All AIPMT

A
$0.8$
B
$0.6$
C
$0.4$
D
$0.2$

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The relative lowering of vapour pressure produced by dissolving $71.5 \ g$ of a substance in $1000 \ g$ of water is $0.00713$. The molecular weight of the substance will be:

Medium

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The partial vapour pressure of any volatile component of a solution is equal to the vapour pressure of the pure component multiplied by its mole fraction in the solution is called:

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At $298 \ K$,$0.714$ moles of liquid $A$ is dissolved in $5.555$ moles of liquid $B$. The vapour pressure of the resultant solution is $475 \ torr$. The vapour pressure of pure liquid $A$ at the same temperature is $280.7 \ torr$. What is the vapour pressure of pure liquid $B$ in $torr$?

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The vapour pressure of pure $CHCl_3$ and $CH_2Cl_2$ are $200 \,atm$ and $41.5 \,atm$ respectively. The weights of $CHCl_3$ and $CH_2Cl_2$ are respectively $11.9 \,g$ and $17 \,g$. The vapour pressure of the solution will be (in $,atm$)

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Which one of the following represents the graph between $\log p$ (on $Y$-axis) and $\frac{1}{T}$ (on $X$-axis)?
($p=$ vapour pressure of a liquid, $T=$ absolute temperature)

MediumTS EAMCET 2003

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The vapour pressure of a solvent decreases by $10 \ mm$ of $Hg$ when a non-volatile solute is added to the solvent. The mole fraction of the solute in the solution is $0.2$. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be $20 \ mm$ of $Hg$?

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The relative lowering of vapour pressure produced by dissolving $71.5 \ g$ of a substance in $1000 \ g$ of water is $0.00713$. The molecular weight of the substance will be:

The partial vapour pressure of any volatile component of a solution is equal to the vapour pressure of the pure component multiplied by its mole fraction in the solution is called:

At $298 \ K$,$0.714$ moles of liquid $A$ is dissolved in $5.555$ moles of liquid $B$. The vapour pressure of the resultant solution is $475 \ torr$. The vapour pressure of pure liquid $A$ at the same temperature is $280.7 \ torr$. What is the vapour pressure of pure liquid $B$ in $torr$?

The vapour pressure of pure $CHCl_3$ and $CH_2Cl_2$ are $200 \,atm$ and $41.5 \,atm$ respectively. The weights of $CHCl_3$ and $CH_2Cl_2$ are respectively $11.9 \,g$ and $17 \,g$. The vapour pressure of the solution will be (in $,atm$)

Which one of the following represents the graph between $\log p$ (on $Y$-axis) and $\frac{1}{T}$ (on $X$-axis)?($p=$ vapour pressure of a liquid, $T=$ absolute temperature)

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Which one of the following represents the graph between $\log p$ (on $Y$-axis) and $\frac{1}{T}$ (on $X$-axis)?
($p=$ vapour pressure of a liquid, $T=$ absolute temperature)