Question

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

• A. 45 $cm^3$
• B. 400 $cm^3$
• C. 450 $cm^3$
• D. 100 $cm^3$

Answer 1

Answer: (B)

400 $cm^3$

Answer 2

$N_1V_1 = N_2 V_2$
$\frac{N}{2} \times V_1 = \frac{N}{10} \times 500$
$V_1$ =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 $cm^3$