One litre solution contains 10.6 gm of  $N{a}_{2}C{O}_3$  and 4 gm NaOH where this solution is treated with HCl in the presence of phenolphthalein indicator consumed 300 ml of HCl. Then the concentration of HCl is M moles/Lit. The value of 10 M. .(rounding to the nearest integer)

Moles of  $N{a}_{2}C{O}_3=\frac{10.6}{106}=0.1$ moles of  $NaOH=\frac{4}{40}=0.1$  in the presence of phenolphthalein HCl neutralizes.  $\frac{1}{2}$  eq of  $N{a}_{2}C{O}_3$  and total NaOH.
0.05 mole of  $N{a}_{2}C{O}_3+0.1$   moles of NaOH = moles of HCl
0.15 moles of HCl needed by 300 ml of mixture
         ?             $\to$                    1000 ml of mixture
$M=\frac{0.15\times 1000}{300}=0.5moles\text{/}Lit$