Let the vapour pressure of pure octane be p₁⁰.

Then, the vapour pressure of the octane after dissolving the non-volatile solute is \(\frac{80}{100}p_1^0\) = 0.8 p₁⁰

Molar mass of solute, M₂ = 40 g mol^-1

Mass of octane, w₁ = 114 g

Molar mass of octane, (C₈H₁₈), M₁ = 8 × 12 + 18 × 1

= 114 g mol^-1

Applying the relation,

\(\frac{p_1^0 - p_1}{p_1^0}\) = \(\frac{w_2 \times M_1}{M_2 \times w_1}\)

⇒ \(\frac{p_1^0 - 0.8p_1^0}{p_1^0}\) = \(\frac{w_2 \times 114}{40 \times 114}\)

⇒ \(\frac{0.2p_1^0}{p_1^0}\) = \(\frac{w_2}{40}\)

⇒ w₂ = 8g

Hence, the required mass of the solute is 8 g.