Chemistry0.4 g mixture of NaOH , Na 2 CO 3 and some inert impurities was first titrated with N 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 Na 2 CO 3 in the mixture is (Rounded off to the nearest integer)

0.4 g mixture of NaOH,Na2CO3 and some inert impurities was first titrated with N10HCl 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)

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    Solution:

    Upto first end point 

     gm equi. of NaOH+Na2CO3=HClx+y×1=110×17.5x+y=1.75 Upto second end point NaOH+Na2CO3HClx+y+2=110×19x+2y=1.9%Na2CO3=0.15×10-3×1060.4×100=3.975%=4%

    Hence answer is (4)

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