Question

$10ml$ of $HCl$ solution gave $0.1435gm$ of $AgCl$ when treated with excess of $AgN{O}_3$. The normality of the $HCl$ solution is:

• A. 0..1 N
• B. 0.01 N
• C. 1 N
• D. 10 N

Answer 1

The correct option is A 0..1 N

$AgN{O}_3+HCl⟶AgCl+HN{O}_3$

$36.5⟶143.5$

$\left(x\right)⟵0.1435$

$x=\frac{0.1435\times 36.5}{143.5}=0.0365$ $grams$

Basicity of $HCl$ is $1$. So Molarity $=$ Normality

$MolarityofHCL=\frac{Givenweight}{GramMolecularweight}\times \frac{1000}{Volume\left(mL\right)}=\frac{0.0365}{36.5}\times 10001=0.1$

Molarity of $HCl=0.1$ $M$

Norrmality of $HCl=0.1$ $N$