Question

Find the normality of the solution prepared by diluting 200 mL of a 1 M $HCl$ with 100 mL of water.

• A. 0.66 N
• B. 1.32 N
• C. 0.33 N
• D. 0.50 N

Answer 1

The correct option is A 0.66 N

n-factor for $HCl=1$,
normality = molarity
$N_1{V}_1=N_2{V}_2$;
final volume will become 300 mL
$(1\times 200)=(N\times 300)$
Normality of the final solution = 0.66 N