A $100 \mathrm{~mL}$ solution was made by adding $1.43 \mathrm{~g}$ of $\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot \mathrm{xH}_{2} \mathrm{O}$. The normality of the solution is $0.1$ N. The value of $x$ is
(The atomic mass of $\mathrm{Na}$ is $23 \mathrm{~g} / \mathrm{mol}$ ) :-
Molar mass of $\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot \mathrm{xH}_{2} \mathrm{O}$
$\Rightarrow 23 \times 2+12+48+18 x$
$\Rightarrow 46+12+48+18 x$
$\Rightarrow(106+18 x)$
Eqwt $=\frac{\mathrm{M}}{2}=(53+9 \mathrm{x})$
As $\mathrm{n}_{\text {factor }}$ in dissolution will be determined from net cationic or anionic charge; which is 2 so
Equt $=\frac{\mathrm{M}}{2}=53+9 \mathrm{x}$
Gmeq $=\frac{w t}{\text { Eqwt }}=\frac{1.43}{53+9 x}$
Normality $=\frac{\text { Gmeq }}{\mathrm{V}_{\text {litre }}}$
Normality $=0.1=\frac{1.43}{\frac{53+9 \mathrm{x}}{0.1}}$
As volume $=100 \mathrm{ml}$
$=0.1$ Litre
$53+9 x=143$
$9 x=90$
$x=10.00$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.