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^-1) of the solute is
[Given that the molar mass of n-octane is 114 g mol^-1.]

• A. $40$
• B. $60$
• C. $80$
• D. $20$

Hint:

Relative lowering of vapour pressure is a colligative property as it depends on the moles of the solute but is independent of its nature.

Answer 1

The Correct Option is A

Let us take a dilute solution,
\(\frac{P_{0}-P_{s}}{P_{s}} \approx\frac{P_{0}-P_{s}}{P_{0}}=\frac{n_{\text{Solute}}}{n_{\text{Solvent}}}\)
Let \(P_0\) = 100,
Vapour Pressure reduced to 80%
\(\therefore P_{s}=80\)

\(\frac{100-80}{100}=\frac{8 m}{114}\)
\(m=40\)

Therefore, the correct option is (A) : 40.

Answer 2

Relative lowering of vapour pressure is a colligative property as it depends on the moles of the solute but is independent of its nature.
Raoult’s law gives the relation between the relative lowering of the vapour pressure of a solution and the mole fraction of the non-volatile solute by:

\(\frac{p_0-p_s}{p^0}=\frac{n_2}{n_1+n_2}\)

• where p₀ denotes the vapour pressure of the pure solvent,
• p_s denotes the vapour pressure of the solution,
• n₂ denotes the number of moles of the solute, and
• n₁ denotes the number of moles of the solvent.

Complete step-by-step solution:

Given: Molar mass of a non-volatile, non-electrolyte solute = 8 grams per mole.

• Mass of octane = 114g
• Also, vapour pressure is reduced to 80 percent.
• The term p₀ - p_s represents the lowering of vapour pressure
• and \(\frac{p_0-p_s}{p^0}\) represent the relative lowering of vapour pressure
• \(\frac{n_2}{n_1+n_2}\) represent the mole fraction of that solute.

Therefore, we can write: \(\frac{p_0-p_s}{p^0}=\frac{\frac{w_2}{M_2}}{\frac{w_2}{M_2}+\frac{w_1}{M_1}}\)

• where w₂ and M₂ represent the mass in grams and molar mass of the solute respectively
• w₁ and M₁ represent the mass in grams and molar mass of the solvent respectively.

Here, we need to find out the mass of the non-volatile, non-electrolyte solute (w₂).
Let the vapour pressure of the solvent (p⁰) be 100.
Since the vapour pressure will be 80% of 100, i.e., 80,
so, \(p_0-p_s=80\)
M₂ = 8 g/mol, w₁ = 114 g
The molar mass of octane, M₁ (C₈H₁₈) = 12 x 8 + 1 x 18
Hence, M₁ = 114 g
Using the formula for the relative lowering of vapour pressure, we can write:

\(\frac{80}{100}=\frac{\frac{w_2}{8}}{\frac{w_2}{8}+\frac{114}{114}}\)

\(⇒0.80=\frac{\frac{w_2}{8}}{\frac{w_2}{8}+1}\)

\(⇒\frac{w_2}{8}+1= \frac{w_2}{8}×0.80\)

\(⇒w_2+8=0.80w_2\)

\(⇒w_2-0.80w_2=8\)

\(⇒0.20w_2=8\)

\(⇒w_2=\frac{8}{0.20}=40g\)

Therefore, the mass of the non-volatile, non-electrolyte solute is 40g.

So, the correct option is (A): \(40\).

Answer 3

Assuming the given solution is dilute.
By putting the values in the formula:

\(\dfrac{\Delta P}{ P^o_A} = \dfrac{n_B}{n_A} = \dfrac{w_B}{m_B} . \dfrac{m_A}{w_A}\)

\(\dfrac{20}{100} = \dfrac{8}{m_b}.\dfrac{114}{114}\)

\(m_B=\dfrac{8X100}{20}\)

\(=40\ gmol^{-1}\)

Hence, the correct answer is Option (A): \(40\)