The volumes of two HCl solutions A (0.5 , M) | vedclass

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(A) Let the volume of solution $A$ be $V_1$ and the volume of solution $B$ be $V_2$.Given $V_1 + V_2 = 2 \, L$,so $V_2 = (2 - V_1) \, L$.Using the mol

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$0.5 \, L$ of $A + 1.5 \, L$ of $B$

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(A) Let the volume of solution $A$ be $V_1$ and the volume of solution $B$ be $V_2$.Given $V_1 + V_2 = 2 \, L$,so $V_2 = (2 - V_1) \, L$.Using the molarity equation $M_1 V_1 + M_2 V_2 = M_3 V_3$:$0.5 	imes V_1 + 0.1 	imes (2 - V_1) = 0.2 	imes 2$$0.5 V_1 + 0.2 - 0.1 V_1 = 0.4$$0.4 V_1 = 0.2$$V_1 = \frac{0.2}{0.4} = 0.5 \, L$Therefore,$V_2 = 2 - 0.5 = 1.5 \, L$.Thus,$0.5 \, L$ of $A$ and $1.5 \, L$ of $B$ are required.

The volumes of two $HCl$ solutions $A$ $(0.5 \, M)$ and $B$ $(0.1 \, M)$ to be mixed for preparing $2 \, L$ of $0.2 \, M \, HCl$ are

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