5 g of was dissolved in deionized water to prepare a 450 mL stock solution. What volume (in mL) of this solution would be required to prepare 500 mL of 0.1 M solution?
Given: Molar Mass of Na, O and H is 23, 16 and 1 g mol respectively
Given: Molar Mass of Na, O and H is 23, 16 and 1 g mol respectively
Show answer and solution
Answer: 180
Solution
Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol.
Moles of NaOH in stock = 5/40 = 0.125 mol.
Concentration of stock = 0.125/0.450 = 5/18 M.
Using M₁V₁ = M₂V₂:
V₁ = (0.1 × 500) / (5/18) = 50 × (18/5) = 180 mL.
Moles of NaOH in stock = 5/40 = 0.125 mol.
Concentration of stock = 0.125/0.450 = 5/18 M.
Using M₁V₁ = M₂V₂:
V₁ = (0.1 × 500) / (5/18) = 50 × (18/5) = 180 mL.
[JEE Main 2023 · 24 Jan · Shift 1 · Paper-1(B.E./B. Tech.)]