Question

The normality of $1.11g$ of $CaC{l}_2$ dissolved in $250mL$ solution is .

• A. $0.1N$
• B. $0.25N$
• C. $0.08N$
• D. $0.3N$

Answer 1

The correct option is C $0.08N$

Normality = $\frac{Gramequivalentsofsolute}{Volumeofsolution\left(inlitre\right)}$

Equivalent mass $=\frac{MolecularMass}{Chargeonion\times numberofions}$
[$Ca\to {Ca}^{2+}+2e^{-}$]
[$C{l}_2+2e^{-}\to 2C{l}^{-}$]
Either way, the denominator would be 2, please check this carefully!

= $\frac{111}{2}$ = 55.5
No of gram equivalents = $\frac{1.11}{55.5}$
$\Rightarrow$ Normality = $\frac{\frac{1.11}{55.5}}{\frac{250}{1000}}=0.08N$