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# One mole of non- volatile solute is dissolved in two moles d water. The vapour pressure of the solution relative to that of water is:A. $\dfrac{2}{3}$B. $\dfrac{1}{3}$ C. $\dfrac{1}{2}$D. $\dfrac{3}{2}$

Last updated date: 01st Mar 2024
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Hint: Vapour pressure is used to measure the tendency of a material to change its state into gaseous state and it increases with the increase in temperature. The formula of relative lowering of vapour pressure is:
$\dfrac{{{P}^{\circ }}-P}{{{P}^{\circ }}}={{x}_{solute}}$
Where, ${{P}^{\circ }}$is the vapour pressure of pure solvent
And P is the vapour pressure of solution.
x is the mole fraction of solute that is the moles of solute divided by the total moles of solute.

Step by step solution:
-We have to give the relation between the vapour pressure of solution to that of water, and one mole of non-volatile solute is given, so this is a case of relative lowering of vapour pressure.
- This question relates to Raoult’s law, according to which vapour pressure of a solution in which non-volatile solute is present, is equal to the vapour pressure of the pure solvent at that temperature and multiplied by its mole fraction.
- So, the formula of relative lowering of vapour pressure is:
$\dfrac{{{P}^{\circ }}-P}{{{P}^{\circ }}}={{x}_{solute}}$
Where, ${{P}^{\circ }}$is the vapour pressure of pure solvent
And P is the vapour pressure of solution.
x is the mole fraction of solute that is the moles of solute divided by the total moles of solute.
- moles of solute given is 1 mole and total moles of solute is 3.
- Now by putting all the values and solving, we get,
$1-\dfrac{P}{{{P}_{\circ }}}=\dfrac{moles\text{ }of\text{ }solute}{total\text{ }moles\text{ }of\text{ }solute}$
$1-\dfrac{P}{{{P}^{\circ }}}=\dfrac{\dfrac{1}{3}}{\dfrac{1}{3}+\dfrac{2}{3}}$

$\Rightarrow 1-\dfrac{1}{3}=\dfrac{2}{3}$
Hence, we can conclude that the correct option is (a) that the vapour pressure of the solution relative to that of water is $\dfrac{2}{3}$.

$Vapour\text{ }pressure\text{ }\propto \text{ }Temperature$