Moles of methanol present in 2.5 L of 0.25 M solution Molarity = $\frac{\text{Moles of }\mathrm{CH_3OH}}{\text{Volume in litre}}$ ⇒ 0.25 = $\frac{\text{Moles of }\mathrm{C_3OH}}{2.5}$ ∴ Moles of $\mathrm{CH_3OH}$ = 2.5 × 0.25 = 0.625 ∴ Mass of $\mathrm{CH_3OH}$ = 0.625 × 32 = 20 g [∴ Molecular mass of $\mathrm{CH_3OH}$ = 12 + 1 × 3 + 16 + 1 = 32] ∴ 0.793 × $10^{3}$ g of $\mathrm{CH_3OH}$ is present in 1000 mL ∴ 20 g of $\mathrm{CH_3OH}$ is present in = $\frac{1000}{0.793 \times 10^{3}}$ × 20 = 25.2 mL