150 mL of 6.00 M solution is mixed with 250 mL of 3.00 M . The resulting molarity of the solution is — Redox Reactions and Volumetric Analysis Chemistry Question
Question
150 mL of 6.00 M $H_2SO_4$ solution is mixed with 250 mL of 3.00 M $H_2SO_4$. The resulting molarity of the solution is:
Answer: A
💡 Solution & Explanation
When mixing two solutions of the same solute, $M_f = \frac{M_1V_1 + M_2V_2}{V_1 + V_2}$. $M_f = \frac{150 \times 6.00 + 250 \times 3.00}{150 + 250} = \frac{900 + 750}{400} = \frac{1650}{400} = 4.125 M$.
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