
The vapour pressure of pure liquid solvent $\,A\,$ is $0.80atm$.When a non- volatile substance $\,B\,$ is added to the solvent its vapour pressure drops to $0.60atm$ .what is the mole fraction of component $\,B\,$ in the solution?
A.$0.25$
B.$0.28$
C.$0.30$
D.$0.31$
Answer
572.1k+ views
Hint: Whenever, a non-volatile solute is added to a solution its vapor pressure gets dropped. Which means, vapor pressure in a solution containing a non-volatile solute is less than that of a pure solvent at any given temperature. So, for this question, the answer can be obtained by applying Raoult’s law.
Formula used: The formula to be used is,
$\Delta p = {p^o}{X_2}$where,
$\Delta p = $change in vapour pressure
${p^o} = $vapour pressure of pure solvent
${X_2} = $mole fraction of solute
Complete step by step answer:
Let us first analyse the given data;
Vapour pressure of pure solvent ${p^o} = 0.80atm$
Changed vapour pressure $p = 0.60atm$
By applying the formula$\Delta p = {p^o}{X_2}$
${X_2} = \dfrac{{\Delta p}}{{{p^o}}}$, substitute the given values, we get,
${X_2} = \dfrac{{0.80 - 0.60}}{{0.80}}$
Which means the answer is;
$ = 0.25$
This is the mole fraction of B. So the correct answer is option A.
Additional information:
Raoult’s law says that the partial pressure of a solvent in a solution is equal to the product of vapour pressure of pure solvent and its mole fraction. By using this it can be said that relative lowering of vapour pressure by the addition of a non-volatile solute will be equal to the products of mole fraction of solute and vapour pressure of solvent.
Lowering of vapour pressure is a colligative property, this lowering by adding a non-volatile solute is because while adding solute to a solvent the solvent evaporates less since the solute occupies the surface which interrupts the vapourization. This makes the amount of vapour less and causes a decrease in vapour pressure.
Note:
Mole fraction doesn’t possess any units since it is the ratio of the number of moles of a component to the total number of moles of all the components in a solution. Total mole fraction of all the components will always be unity, that is one.
Formula used: The formula to be used is,
$\Delta p = {p^o}{X_2}$where,
$\Delta p = $change in vapour pressure
${p^o} = $vapour pressure of pure solvent
${X_2} = $mole fraction of solute
Complete step by step answer:
Let us first analyse the given data;
Vapour pressure of pure solvent ${p^o} = 0.80atm$
Changed vapour pressure $p = 0.60atm$
By applying the formula$\Delta p = {p^o}{X_2}$
${X_2} = \dfrac{{\Delta p}}{{{p^o}}}$, substitute the given values, we get,
${X_2} = \dfrac{{0.80 - 0.60}}{{0.80}}$
Which means the answer is;
$ = 0.25$
This is the mole fraction of B. So the correct answer is option A.
Additional information:
Raoult’s law says that the partial pressure of a solvent in a solution is equal to the product of vapour pressure of pure solvent and its mole fraction. By using this it can be said that relative lowering of vapour pressure by the addition of a non-volatile solute will be equal to the products of mole fraction of solute and vapour pressure of solvent.
Lowering of vapour pressure is a colligative property, this lowering by adding a non-volatile solute is because while adding solute to a solvent the solvent evaporates less since the solute occupies the surface which interrupts the vapourization. This makes the amount of vapour less and causes a decrease in vapour pressure.
Note:
Mole fraction doesn’t possess any units since it is the ratio of the number of moles of a component to the total number of moles of all the components in a solution. Total mole fraction of all the components will always be unity, that is one.
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