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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. $\dfrac{3}{2}$

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Last updated date: 01st Mar 2024
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IVSAT 2024
Answer
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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 $\dfrac{2}{3}$.

Additional information:
We will now see how some factors affects the vapour pressure:
- It is observed that as the temperature of the liquid is increased then the vapour pressure also increases.
\[Vapour\text{ }pressure\text{ }\propto \text{ }Temperature\]
- We can also see that the vapour pressure is hardly influenced by any changes in pressure.

Note
- The given question is an example of an ideal solution, that is the solution which obeys Raoult’s law. There are several other examples of ideal solutions that obeys this law are Toluene and Benzene, Ethyl iodide and Ethyl Bromide.