(A) According to Raoult's law,the relative lowering of vapour pressure is equal to the mole fraction of the solute.$\frac{p - p_s}{p} = X_{solute}$Sin

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$\frac{p - p_s}{p} = \frac{n}{n + N}$

(A) According to Raoult's law,the relative lowering of vapour pressure is equal to the mole fraction of the solute.$\frac{p - p_s}{p} = X_{solute}$Since the mole fraction of the solute $X_{solute}$ is given by $\frac{n}{n + N}$,where $n$ is the number of moles of the solute and $N$ is the number of moles of the solvent.Therefore,the expression is $\frac{p - p_s}{p} = \frac{n}{n + N}$.Thus,the correct answer is $A$.

Class 12 Chemistry Solutions Lowering of vapour pressure

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Which one of the following is the expression of Raoult's law?
$p$ = vapour pressure of pure solvent
$p_s$ = vapour pressure of the solution
$n$ = number of moles of the solute
$N$ = number of moles of the solvent

A
$\frac{p - p_s}{p} = \frac{n}{n + N}$
B
$\frac{p_s - p}{p} = \frac{N}{N + n}$
C
$\frac{p - p_s}{p_s} = \frac{N}{N - n}$
D
$\frac{p_s - p}{p_s} = \frac{N - n}{N}$

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Lowering of vapour pressure

$60 \ g$ of Urea (Mol. wt $60$) was dissolved in $9.9 \ moles$ of water. If the vapour pressure of pure water is $P_o$,the vapour pressure of the solution is: (in $P_o$)

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In $108 \, g$ of water,$18 \, g$ of a non-volatile compound is dissolved. At $100^{\circ} C$,the vapour pressure of the solution is $750 \, mm \, Hg$. Assuming that the compound does not undergo association or dissociation,the molar mass of the compound in $g \, mol^{-1}$ is

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Two liquids $A$ and $B$ have $P_A^o$ and $P_B^o$ in the ratio of $1 : 3$ and the ratio of number of moles of $A$ and $B$ in the liquid phase is $1 : 3$. The mole fraction of $A$ in the vapour phase in equilibrium with the solution is equal to:

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The vapour pressure of water at room temperature is $23.8 \ mm \ Hg$. The vapour pressure of an aqueous solution of sucrose with mole fraction of $0.1$ is equal to .......... $mm \ Hg$.

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$2\, \text{moles}$ of liquid $A$ is mixed with $3\, \text{moles}$ of liquid $B$. Calculate the mole fraction of $A$ in the vapour phase if $P_A^o = 100\, \text{mm}$ and $P_B^o = 200\, \text{mm}$.

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Which one of the following is the expression of Raoult's law?$p$ = vapour pressure of pure solvent$p_s$ = vapour pressure of the solution$n$ = number of moles of the solute$N$ = number of moles of the solvent

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$60 \ g$ of Urea (Mol. wt $60$) was dissolved in $9.9 \ moles$ of water. If the vapour pressure of pure water is $P_o$,the vapour pressure of the solution is: (in $P_o$)

In $108 \, g$ of water,$18 \, g$ of a non-volatile compound is dissolved. At $100^{\circ} C$,the vapour pressure of the solution is $750 \, mm \, Hg$. Assuming that the compound does not undergo association or dissociation,the molar mass of the compound in $g \, mol^{-1}$ is

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Which one of the following is the expression of Raoult's law?
$p$ = vapour pressure of pure solvent
$p_s$ = vapour pressure of the solution
$n$ = number of moles of the solute
$N$ = number of moles of the solvent