When 5 , mL of N , HCl,20 , mL of N/2 , H_2SO | vedclass

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(D) The total number of milliequivalents $(N_1V_1 + N_2V_2 + N_3V_3)$ is calculated as follows: $N_1V_1 = 5 \, mL 	imes 1 \, N = 5 \, meq$ $N_2V_2 =

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$N/40$

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(D) The total number of milliequivalents $(N_1V_1 + N_2V_2 + N_3V_3)$ is calculated as follows: $N_1V_1 = 5 \, mL 	imes 1 \, N = 5 \, meq$ $N_2V_2 = 20 \, mL 	imes 0.5 \, N = 10 \, meq$ $N_3V_3 = 30 \, mL 	imes (1/3) \, N = 10 \, meq$ Total milliequivalents = $5 + 10 + 10 = 25 \, meq$. The final volume of the solution is $1 \, L = 1000 \, mL$. Normality $(N)$ = $\frac{	ext{Total milliequivalents}}{	ext{Total volume in mL}} = \frac{25}{1000} = \frac{1}{40} \, N$ or $N/40$.

When $5 \, mL$ of $N \, HCl$,$20 \, mL$ of $N/2 \, H_2SO_4$,and $30 \, mL$ of $N/3 \, HNO_3$ are mixed together and the volume is made up to $1 \, L$,the normality of the resulting solution is:

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