The amount of ${\text{Na}}_2{\text{CO}}_3$   present in 100 mL of 0.1M ${\text{Na}}_2{\text{CO}}_3$  solution is

Concentration of${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$  $=0.1\text{M}$

GMW of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}=106\text{ g/mol}$

Weight of ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}=?$

From the formula of molarity,

$\text{M}=\frac{\text{w}}{\text{GMW}}\times \frac{1000}{\text{V in mL}}$

$\Rightarrow \text{w}=\frac{\text{M×GMW}}{1000}\times \text{V}\text{in mL}$

Substitute all known data in the above equation to find weight of${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ .

$\text{Weight}{\text{ of Na}}_{\text{2}}{\text{CO}}_{\text{3}},\text{w}=\frac{0.1\times 106\times 100}{1000}$

                               $\therefore w=1.06g$