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Last updated date: 01st Dec 2023
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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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Hint: We have to know that, the endless supply of a solute, the fume pressing factor of the arrangement will diminish. To address this, expect the mass of the solute to be x and track down the quantity of moles of the solute just as octane. At that point utilize Raoult's law as per which, the overall bringing down of fume pressure is corresponding to the mole part of the solute.

We need to see, fume pressure is the pressing factor applied by the gas on the fluid in a closed system at a specific temperature. It is the vertical pressing factor, that the atoms of the fluid face, when they are attempting to leave the outside of the fluid through vanishing. In straightforward words, we can say that it is the pressing factor, which is displayed by the fume present over the fluid surface. Bringing down a fume endless supply of solute is given by Raoult's law. Presently let us talk about the given inquiry, here let us expect that the necessary mass of the non-unpredictable solute is $x$ gram. Hence, the quantity of moles of solute will be,
$\dfrac{x}{{40}} = 0.025x{\text{ }}moles$ .
Where, the molar mass of the solute = $40g/mol$ .
The mass of octane = $114g$
The molar mass of octane = $114g/mol$ .
Thus, that the quantity of moles of octane is one mole.
As per Raoult's law, the general bringing down of fume pressure is relative to the mole part of the solute.
The mole fraction of solute = $\dfrac{{0.025x}}{{1 + 0.025x}}$
Then, the vapor pressure is lowered by $20\%$ ,
$\dfrac{{0.025x}}{{1 + 0.025x}} = \dfrac{{20}}{{100}}$
Rewrite the above expression, to find out the $x$ ,
$100 \times 0.025x = 20\left( {1 + 0.025x} \right)$
Then,
$2.5x = 20 + 0.5x$
Next, we have to write,
$2x = 20$
Therefore,
$x = 10$
Hence, $x$ gram = $10g$ .

The correct option is (A).

Note:
At the point when the fume pressure changed because of the expansion of the solute. This bringing down in fume pressure is a colligative property of a fluid. The property of an answer that changes upon the expansion of solute is known as the colligative properties.