A 100mL solution was made by adding


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

$53+9 x=143$

$9 x=90$


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