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)
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
answer is 4.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Upto first end point gm equi. of NaOH+Na2CO3=HClx+y×1=110×17.5x+y=1.75 Upto second end point NaOH+Na2CO3≡HClx+y+2=110×19x+2y=1.9%Na2CO3=0.15×10-3×1060.4×100=3.975%=4%Hence answer is (4)
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)