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The mass of a non-volatile solute of molar mass 40g/mol that should be dissolved in 114g of octane to lower its vapour pressure by 20% is:
[A] 10g
[B] 11.4g
[C] 9.8g
[D] 12.8g

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Last updated date: 20th Jun 2024
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Answer
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Hint: Upon addition of a solute, the vapour pressure of the solution will decrease. To solve this, assume the mass of the solute to be x and find the number of moles of the solute as well as octane. Then use Raoult’s law according to which, the relative lowering of vapour pressure is proportional to the mole fraction of the solute.

Complete step by step answer:
We know that, vapour pressure is the pressure exerted by the gas on liquid (or solid) in a closed system at a particular temperature. It is the upward pressure that the molecules of the liquid face when they are trying to leave the surface of the liquid through evaporation. In simple words, we can say that it is the pressure which is exhibited by the vapour present above the liquid surface.
Lowering of vapour pressure upon addition of a solute is given by Raoult’s law.

Now let us discuss the given question.
Here, let us assume that the required mass of the non-volatile solute is ‘x’ g.
The molar mass of the solute is 40 g/mol.
Therefore, the number of moles of solute will be $\dfrac{x}{40}$ = 0.025x moles.
Now, mass of octane = 114 g and molar mass of octane = 114 g/mol.
Therefore, the number of moles of octane = 1 mole.
Now, according to Raoult’s law, the relative lowering of vapour pressure is proportional to the mole fraction of the solute.
So, mole fraction of solute is $\dfrac{0.025x}{1+0.025x}$ .

Now, the vapour pressure is to be lowered by 20%.
So, we can write that $\dfrac{0.025x}{1+0.025x}=\dfrac{20}{100}$
Now, solving for x, we will get-
     $\begin{align}
 & 100\times 0.025x=20\left( 1+0.025x \right) \\
 & 0r,2.5x=20+0.5x \\
 & or,2x=20 \\
 & or,x=10 \\
\end{align}$
We can see from the above calculation that the mass of a non-volatile solute that should be dissolved in 114g of octane to lower its vapour pressure by 20% is 10g.
So, the correct answer is “Option A”.

Note: Here, the vapour pressure changed due to addition of the solute. This lowering in vapour pressure is a colligative property of a liquid. The property of solution that changes upon addition of solute is known as the colligative properties.
There are 4 colligative properties shown by liquids. They are-osmotic pressure, depression of freezing point, elevation of boiling point and relative lowering of vapour pressure.