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(C) For $HCl$,the normality $(N)$ is equal to the molarity $(M)$ because the n-factor is $1$. Thus,$2 \ N = 2 \ M$ and $5 \ N = 5 \ M$.Let $V_1$ be th

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$750$

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(C) For $HCl$,the normality $(N)$ is equal to the molarity $(M)$ because the n-factor is $1$. Thus,$2 \ N = 2 \ M$ and $5 \ N = 5 \ M$.Let $V_1$ be the volume of $2 \ M \ HCl$ used and $V_2$ be the volume of $5 \ M \ HCl$ used.We are given $V_1 = 500 \ mL$ and we have a maximum of $500 \ mL$ of $5 \ M \ HCl$ available $(V_2 \leq 500 \ mL)$.The mixture equation is $M_1 V_1 + M_2 V_2 = M_{final} (V_1 + V_2)$.Substituting the values: $2(500) + 5(V_2) = 3(500 + V_2)$.$1000 + 5V_2 = 1500 + 3V_2$.$2V_2 = 500$,which gives $V_2 = 250 \ mL$.Since $250 \ mL The total volume of $3 \ M \ HCl$ prepared is $V_1 + V_2 = 500 \ mL + 250 \ mL = 750 \ mL$.

You are supplied with $500 \ mL$ each of $2 \ N \ HCl$ and $5 \ N \ HCl$. What is the maximum volume of $3 \ M \ HCl$ that you can prepare using only these two solutions (in $mL$)?

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