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  4. The volume of water to be added to N/2 HCl to prepare 500 cm3 of N/10 solution is

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Q. The volume of water to be added to N/2 HCl to prepare $500 cm^3$ of N/10 solution is

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Solution:

$N_1V_1 = N_2 V_2$
$\frac{N}{2} \times V_1 = \frac{N}{10} \times 500$
$V_1$ =100
Vol. of N/10 solution = 500 $cm^3$
Vol. of N/2 solution = 100 $cm^3$
$\therefore $ Water to be added = (500 - 100) $cm^3$ = 400 $cm^3$