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).

Chemistry · Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

JEE Main 2025 — 22 January, Evening Shift — Question 47

$20\ \mathrm{mL}$ of $2\ \mathrm{M}$ $\mathrm{NaOH}$ solution is added to $400\ \mathrm{mL}$ of $0.5\ \mathrm{M}$ $\mathrm{NaOH}$ solution. The final concentration of the solution is $\_\_\_\_ \times 10^{-2}\ \mathrm{M}$ (nearest integer).

Answer: 57.1

Numerical answer — enter this value.

Step-by-step solution

Moles of $\mathrm{NaOH}$ from first solution $\mathrm{= 0.020 \times 2 = 0.040\ mol}$

Moles of $\mathrm{NaOH}$ from second solution $\mathrm{= 0.400 \times 0.5 = 0.200\ mol}$

Total moles of $\mathrm{NaOH}$ $\mathrm{= 0.040 + 0.200 = 0.240\ mol}$

Total volume $\mathrm{= 20 + 400 = 420\ mL = 0.420\ L}$

Final molarity, $\mathrm{M = \dfrac{0.240}{0.420} = 0.571\ M}$

Expressing in required form, we have

\mathrm{0.571\ M = 57.1 \times 10^{-2}\ M}

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Exam: JEE Main 2025
Paper: 22 January, Evening Shift
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry
Topic: Concentration Terms and Their Interconversion

Density of 3\ M NaCl solution is 1.25\ g\ mL^-1 . The molality of the solution is:

Answer: 57.1

Step-by-step solution

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