Question

A solution is formed by diluting $250mL$ of $0.400N{H}_{2}S{O}_4$ with one litre of water. The normality of above formed solution is :

• A. 0.400 N
• B. 0.899 N
• C. 0.040 N
• D. 0.080 N

Answer 1

Answer: (D)

According to law of dilution
$N_1{V}_1=N_2{V}_2$
(equivalent before and after dilution remains same)
$V_1=250mL{V}_2=1000+250=1250mL$
$0.4\times 250=1250\times N_2$
$\frac{0.4\times 250}{1250}=N_2$
$N_2=0.08N$
Normality of a solution is defined as the number of gram equivalent of solute present in one litre of the solution or milii equivalents of solute present in one milli litre of the solution
$N=\frac{\text{ gram equivalent of solute }}{\text{ vol. of sol. (L) }}$
$N=\frac{\text{ mass of solute }}{\text{ equivalent wt. of solute molecule }}÷\text{ vol of solution }(L)$
Solution may be normal $1N$, decinormal $(1/10N)$, centinormal $(1/100N)$ etc.