The mass of non-volatile, non-electrolyte solute (molar mass 50 g/mol) needed to be dissolved in 114 g octane to reduce its vapour pressure to 75%

Octane has a molar mass of 114 g/mol.
ΔP/P = W₂M₁/(W₂M₁ + W₁M₂)
75/100 = W₂(50 g/mol) / (W₂(50 g/mol) + 114 g (114 g/mol))
0.75 = W₂(50) / (W₂(50) + 114(114))
0.75 = W₂(50) / (50W₂ + 12996)
50W₂ + 12996 = 50W₂/0.75
50W₂ + 12996 = 66.67W₂
12996 = 16.67W₂
W₂ = 12996 / 16.67
W₂ = 779.4 g ≈ 780 g

Note: There's a calculation error in the provided solution. The correct approach uses Raoult's Law: ΔP/P = X₂ where X₂ is the mole fraction of the solute. The corrected calculation is shown above.