JEE Main 2025 Jan 22 Shift 2, Chemistry Q25: Types Concentration And Solubility of Solutions

JEE Main2025Jan 22, Shift 2Chemistry

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 ______ $\times 1 0^{- 2}$ M. (Nearest integer).

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Solution

For mixing two solutions of the same solute, the final molarity is the total moles divided by total volume.

Moles in part 1: $M_{1} V_{1} = 2 \times 20 = 40 mmol$ .

Moles in part 2: $M_{2} V_{2} = 0.5 \times 400 = 200 mmol$ .

Total volume: $V_{1} + V_{2} = 20 + 400 = 420 mL$ .

$M_{F} = \frac{40 + 200}{420} = \frac{240}{420} = 0.5714 M = 57.14 \times 1 0^{- 2} M$ .

Nearest integer: $57$ .

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JEE Main 2025 Jan 22 Shift 2, Chemistry Q25: Types Concentration And Solubility of Solutions

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