Question

In a normal solution of ${BaCl}_2$, normalities of ${Ba}^{2+}$ and ${Cl}^{-}$are in the ratio :

• A. $2:1$
• B. $1:2$
• C. $1:1$
• D. $2:3$

Answer 1

The correct option is C $1:1$

Let m be the moles of $BaC{l}_2$.

Moles of $B{a}^{2+}=mmoles$, valency factor of $B{a}^{2+}=2$

Moles of $C{l}^{-}=2mmoles$, valency factor of $C{l}^{-}=1$

Now, equivalents $=moles\times v.f.$

Therefore, $B{a}^{2+}$ equivalents$=C{l}^{-}$equivalents $=2m$

Hence, the ratio is $1:1$.