10 ml of $NaHC{O}_3$ and $N{a}_{2}C{O}_3$ solution requires 2 ml of 0.1 M $H_{2}S{O}_4$ using phenolphthalein as indicator. Weight of $N{a}_{2}C{O}_3$ in mixture is (in gr)

To neutralize whole $N{a}_{2}C{O}_3$, Volume of $H_{2}S{O}_4$ will be twice of given volume. $N{a}_{2}C{O}_3+H^{+}\to NaHC{O}_3+N{a}^{+}$

$$\begin{gathered} N_1{V}_1=N_2{V}_2 \\ {N}_1=\frac{N_2{V}_2}{V_1} \\ {N}_1=\frac{(2\times 0.1)\times (2\times 2)}{10}=0.08N \end{gathered}$$

Weight of $N{a}_{2}C{O}_3$ in 1 $L=0.08\times 53$  $=4.24g$
 Ln 1 L weight of  $N{a}_{2}C{O}_3=4.24g$
Ln 10 ml $N{a}_{2}C{O}_3$ will be  $=\frac{4.24}{1000}\times 10$  $=0.0424g$