Three solutions of HCl having normalities 12 | vedclass

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(D) The condition for mixing solutions is given by the equation: $N_{1}V_{1} + N_{2}V_{2} + N_{3}V_{3} = N_{R}(V_{1} + V_{2} + V_{3})$.Given $N_{1} =

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$1 : 2 : 4$

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(D) The condition for mixing solutions is given by the equation: $N_{1}V_{1} + N_{2}V_{2} + N_{3}V_{3} = N_{R}(V_{1} + V_{2} + V_{3})$.Given $N_{1} = 12$,$N_{2} = 6$,$N_{3} = 2$,and the resultant normality $N_{R} = 4$.Substituting the values: $12V_{1} + 6V_{2} + 2V_{3} = 4(V_{1} + V_{2} + V_{3})$.Simplifying: $12V_{1} + 6V_{2} + 2V_{3} = 4V_{1} + 4V_{2} + 4V_{3}$.$8V_{1} + 2V_{2} = 2V_{3}$,which simplifies to $4V_{1} + V_{2} = V_{3}$.Checking the options:For $A$ $(1:1:5)$: $4(1) + 1 = 5$. This is correct.For $B$ $(1:2:6)$: $4(1) + 2 = 6$. This is correct.For $C$ $(2:1:9)$: $4(2) + 1 = 9$. This is correct.For $D$ $(1:2:4)$: $4(1) + 2 = 6 
eq 4$. This is incorrect.

Three solutions of $HCl$ having normalities $12 \ N$,$6 \ N$,and $2 \ N$ are mixed to obtain a solution of $4 \ N$ normality. Which among the following volume ratios is incorrect for the above mixture?

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