Question

If 8 g of a non-electrolyte solute is dissolved in
114
g
of n-octane to reduce its vapour pressure to
80%, the molar mass (in g mol-") of the solute is
[Given that molar mass of n-octane is 114 g mol-1]
[NEET-2020 (Phase-2)]
(1) 20
(2) 40
(3) 60
(4) 80​

Answer 1

by using this formula you can get your answer... hope you find helpfull

Answer 2

Concept:

Nonelectrolytes are often defined as substances that don't have any distinct ionic form to exist in once they are dissolved in an solution.

Given:

We are given a $8g$ of a non-electrolyte solute is dissolved in $114g$ of $n-$octane to scale back its force per unit area to$80\%$.

Find:

we have to seek out the molar mass of the solute.

Solution:

We will solve this by using Raoult's Law.

As we know, Raoult’s law states that a solvent’s partial pressure in a very solution (or mixture) is equal or a twin of the pressure of the pure solvent multiplied by its mole fraction within the solution.

Here, we'll use the formula $\frac{\Delta P}{P_{A}^0}=\frac{n_{B}}{n_{A}}=\frac{w_{B}}{m_{B}}\cdot\frac{m_{A}}{w_{A}}$ where $\Delta P=100-80=20\%$, $P_{A}^0=100$ , $w_{B}=8$ and $m_{A}=w_{A}=114$.

Now, we'll substitute these values within the formula, we get

$\begin{aligned}\frac{20}{100}&=\frac{8}{m_{B}}\times \frac{114}{114}\\ \frac{2}{10}&=\frac{8}{m_{B}}\end$

Further, we are going to find the worth of $m_{B},$ we get

$\begin{aligned}m_{B}&=\frac{8\times 10}{2}\\ &=\frac{80}{2}\\ &=40\end$

Hence, the molar mass of solute is $40$. So, option (2) is correct.

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