Question

$5.3g$ of $M_{2}C{O}_3$ is dissolved in $150mL$ of $1NHCl$, the unused acid required $100mL$ of $0.5NNaOH$. Hence equivalent weight of $M$ is:

• A. 23
• B. 12
• C. 24
• D. 13

Answer 1

The correct option is A 23

$\text{Milliequivalents of Acid used = Milliequivalents of base taken}$

Milliequivalent of acid used
$=150\times 1-100\times 0.5=100meq$
$\text{Equivalent of acid used = 0.1 equivalents}$
$\text{Equivalents of}{M}_{2}C{O}_3\text{= Equivalents of Acid used = 0.1 equivalents}$

$\text{Equivalent weight of}{M}_{2}C{O}_3\text{(by data) = Equivalent weight of}{M}_{2}C{O}_3$ (by formula)

$\frac{\text{given weight}}{\text{no. of equivalents}}=\frac{\text{Molar weight}}{\text{n-factor}}=\text{Equivalent weight}$

Let '$x$' be the atomic weight of M
$\Rightarrow$$\frac{5.3}{0.1}=\frac{2x+60}{2}$
$x=23$

Equivalent weight of $M=\frac{23}{1}=23$
as $M$ is present as $M^{+}$ in $M_{2}C{O}_3,$ its n-factor is $1$.