250 , mL,6 , M HCl and 350 , mL,8 , M HNO_3 a | vedclass

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(A) For $HCl$,normality $(N)$ = molarity $(M)$ $	imes$ n-factor = $6 	imes 1 = 6 \, N$.For $HNO_3$,normality $(N)$ = molarity $(M)$ $	imes$ n-facto

Vedclass · Accepted Answer

$833.3$

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(A) For $HCl$,normality $(N)$ = molarity $(M)$ $	imes$ n-factor = $6 	imes 1 = 6 \, N$.For $HNO_3$,normality $(N)$ = molarity $(M)$ $	imes$ n-factor = $8 	imes 1 = 8 \, N$.Total milliequivalents of acid = $(250 	imes 6) + (350 	imes 8) = 1500 + 2800 = 4300 \, meq$.Let the volume of water added be $x \, mL$.The total volume of the final solution = $(250 + 350 + x) \, mL = (600 + x) \, mL$.Given the final concentration is $3 \, N$,we use the formula $N_1V_1 + N_2V_2 = N_{final}V_{final}$.$4300 = 3 	imes (600 + x)$.$4300 = 1800 + 3x$.$3x = 2500$.$x = 833.3 \, mL$.

$250 \, mL$,$6 \, M$ $HCl$ and $350 \, mL$,$8 \, M$ $HNO_3$ are mixed with water to make the strength of the resulting solution $3 \, N$. What is the volume of water added? $mL$

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