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

Question

The volume of water to be added to N/2 HCl to prepare 500 cm3 of N/10 solution is (A) 45 cm3 (B) 400 cm3 (C) 450 cm3 (D) 100 cm3

Tardigrade Admin · Accepted Answer

N1V1 = N2 V2 (N/2) × V1 = (N/10) × 500 V1 =100 Vol. of N/10 solution = 500 cm3 Vol. of N/2 solution = 100 cm3 ∴ Water to be added = (500 - 100) cm3 = 400 cm3

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

45 $c m^{3}$

400 $c m^{3}$

450 $c m^{3}$

100 $c m^{3}$

$N_{1} V_{1} = N_{2} V_{2}$
$\frac{N}{2} \times V_{1} = \frac{N}{10} \times 500$
$V_{1}$ =100
Vol. of N/10 solution = 500 $c m^{3}$
Vol. of N/2 solution = 100 $c m^{3}$
$∴$ Water to be added = (500 - 100) $c m^{3}$ = 400 $c m^{3}$

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

45 cm3

400 cm3

450 cm3

100 cm3