JEE Main 2024 Chemistry question
Volume of 3 M NaOH (formula weight 40 g mol⁻¹) which can be prepared from 84 g of NaOH is × 10⁻¹ dm³.
Answer: 7
Solution
Moles of NaOH = 84/40 = 2.1. Volume = moles/molarity = 2.1/3 = 0.7 dm³ = 7 × 10⁻¹ dm³.
Volume of 3 M NaOH (formula weight 40 g mol⁻¹) which can be prepared from 84 g of NaOH is × 10⁻¹ dm³.
Moles of NaOH = 84/40 = 2.1. Volume = moles/molarity = 2.1/3 = 0.7 dm³ = 7 × 10⁻¹ dm³.