Q. 20 mL of 2 M NaOH solution is added to 400 mL of 0.5 M NaOH solution.
The final concentration of the solution is ______ × 10−2 M.
(Nearest integer)
Step 1: Moles of NaOH in first solution
= 2 × (20/1000) = 0.04 mol
Step 2: Moles of NaOH in second solution
= 0.5 × (400/1000) = 0.20 mol
Step 3: Total moles after mixing
= 0.04 + 0.20 = 0.24 mol
Step 4: Total volume
= 20 + 400 = 420 mL = 0.42 L
Step 5: Final molarity
= 0.24 / 0.42 = 0.571 M
Step 6: Required form
0.571 M = 57 × 10−2 M
✔ Final Answer: 57