JEE Main 2024 Chemistry question
From 6.55 g of aniline, the maximum amount of acetanilide that can be prepared will be× 10 g.
Answer: 95
Solution
6.55/93 = 0.0704 mol aniline gives 0.0704 mol × 135 g mol⁻¹ = 9.5 g = 95 × 10⁻¹ g.
From 6.55 g of aniline, the maximum amount of acetanilide that can be prepared will be× 10 g.
6.55/93 = 0.0704 mol aniline gives 0.0704 mol × 135 g mol⁻¹ = 9.5 g = 95 × 10⁻¹ g.