Methylation of 10 g of benzene gave 9.2 g of toluene. Calculate the percentage yield of toluene (Nearest integer)

The methylation of benzene produces toluene according to the Friedel-Crafts alkylation reaction: $$\text{C}_6\text{H}_6 + \text{CH}_3\text{Cl} \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{CH}_3 + \text{HCl}$$.

The molar mass of benzene is $$78 \text{ g/mol}$$ and that of toluene is $$92 \text{ g/mol}$$. Starting with 10 g of benzene, the moles of benzene are $$\frac{10}{78} = 0.1282 \text{ mol}$$. Since the stoichiometry is 1:1, the theoretical yield of toluene is $$0.1282 \times 92 = 11.795 \text{ g}$$.

The actual yield is 9.2 g. Therefore, the percentage yield is $$\frac{9.2}{11.795} \times 100 = 78.0\%$$.

The answer is $$78$$.