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The volumes of two $H C l$ solutions A (0.5M) and B (0.1M) to be mixed for preparing 2 L of 0.2 M $H C l$ are

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1.5 L of A + 0.5 L of B

0.75 L of A + 1.25 L of B

0.5 L of A + 1.5 L of B

1 L of A + 1 L of B

answer is A.

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Detailed Solution

$M = \frac{M_{1} V_{1} + M_{2} V_{2}}{V}$

0.2 = $\frac{0.5 V_{1} + 0.1 (2 - V_{1})}{2}$

on solving V₁ = 0.5 lit, V₂ = 1.5 lit

V₁ = volume of (A) HCl solution

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