0.4 g mixture of NaOH, Na₂CO₃ and some inert impurities was first titrated with ${N \over {10}}$ HCl using phenolphthalein as an indicator, 17.5 mL of HCl was required at the end point. After this methyl orange was added and titrated. 1.5 mL of same HCl was required for the next end point. The weight percentage of Na₂CO₃ in the mixture is ________. (Rounded off to the nearest integer)

<b>1^st end point reaction :</b><br><br>$NaOH + HCl\buildrel {} \over \longrightarrow NaCl + {H_2}O$<br><br>$nf = 1$<br><br>$NaC{O_3} + HCl\buildrel {} \over \longrightarrow NaHC{O_3}$<br><br>$nf = 1$<br><br>Eq of HCl used $= {n_{NaOH}} \times 1 + {n_{N{a_2}C{O_3}}} \times 1$<br><br>$17.5 \times {1 \over {10}} \times {10^{ - 3}} = {n_{NaOH}} + {n_{N{a_2}C{O_3}}}$<br><br><b>2^nd end point :</b><br><br>$NaHC{O_3} + HCl\buildrel {} \over \longrightarrow {H_2}C{O_3}$<br><br>$$1.5 \times {1 \over {10}} \times {10^{ - 3}} = {n_{NaHC{O_3}}} \times 1 = {n_{NaHC{O_3}}}$$<br><br>0.15 mmol = ${n_{N{a_2}C{O_3}}}$<br><br>$0.15 = {n_{N{a_2}C{O_3}}}$<br><br>$${w_{N{a_2}C{O_3}}} = {{0.15 \times 106 \times {{10}^{ - 3}}} \over {0.4}} \times 100 \times 10$$<br><br>= 3.975% <br><br>$\simeq$ 4%