Question

The amount of sodium carbonate to prepare
100 ml of 0.1N solution is​

Answer 1

Answer:

0.53 g

Explanation:

Given,

100ml of 0.1N(normal) of sodium carbonate(Na2Co3)

Normality=0.1

Molecular weight =106

$<strong>equivalent</strong><strong> </strong><strong>weight</strong><strong>=</strong><strong>\</strong><strong>frac</strong><strong>{</strong><strong>Molecular</strong><strong> </strong><strong>weight</strong><strong>}</strong><strong>{</strong><strong>valence</strong><strong> </strong><strong>factor</strong><strong>}</strong>$

Valence factor=2

we know that,

$\bold{\boxed{Normality=\frac{weight}{equivalent weight}×\frac{1000}{Vml}}}$

$0.1=\frac{wt.}{\frac{106}{2}}×\frac{1000}{100}$

$0.1=\frac{wt.}{53}×10$

$wt.=0.1×\frac{53}{10}$

$wt.=0.53$