50 mL of 10 N H_2SO_4,25 mL of 12 N HCl | vedclass

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(A) The total number of equivalents in the mixture is given by the sum of equivalents of individual acids: $N_1V_1 + N_2V_2 + N_3V_3 = N_{resultant} \

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$1$

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(A) The total number of equivalents in the mixture is given by the sum of equivalents of individual acids: $N_1V_1 + N_2V_2 + N_3V_3 = N_{resultant} 	imes V_{total}$.Given:$N_1V_1 = 50 \ mL 	imes 10 \ N = 500 \ meq$$N_2V_2 = 25 \ mL 	imes 12 \ N = 300 \ meq$$N_3V_3 = 40 \ mL 	imes 5 \ N = 200 \ meq$Total equivalents = $500 + 300 + 200 = 1000 \ meq$.Final volume of the mixture = $1000 \ mL$.Normality of the resultant solution $(N)$ = $\frac{	ext{Total equivalents}}{	ext{Total volume in mL}} = \frac{1000 \ meq}{1000 \ mL} = 1 \ N$.

$50 \ mL$ of $10 \ N \ H_2SO_4$,$25 \ mL$ of $12 \ N \ HCl$,and $40 \ mL$ of $5 \ N \ HNO_3$ were mixed together,and the volume of the mixture was made $1000 \ mL$ by adding water. The normality of the resultant solution will be.....$N$.

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