Q21 / 90 · JEE Main 2021 (22 Jul Shift 1) · Chemistry

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

Correct answer: 78

Solution

$C_{6} H_{6} + {CH}_{3} Cl ⟶ C_{6} H_{5} {CH}_{3} + HCl 10 gm (\frac{10}{78} \times 92) gm$
Percentage yield of toluene = $\frac{Actual yield}{Theoretical yield} \times 100$

$= \frac{9.2}{920} \times 78 \times 100 = 78 %$

Hence, the correct answer is 78.

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

Solution

Relevant Questions

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