Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
19 g of a mixture containing NaHCO3 and Na2CO3 on complete heating liberated 1.12 L of CO2 at STP. The weight of the remaining solid was 15.9 g. What is the weight (in g) of Na2CO3 in the mixture before heating?
ChemistrySome Basic Concepts of ChemistryTS EAMCETTS EAMCET 2011
Options:
  • A 8.4
  • B 15.9
  • C 4.0
  • D 10.6
Solution:
1139 Upvotes Verified Answer
The correct answer is: 10.6
Molecular weight of
NaHCO3=23+1+12+48=84
Molecular weight of
Na2CO3=46+12+48=106
Hence, total weight =84+106=190
In 190 g of a mixture, weight of Na2CO3 is
=106
In 19 g of a mixture weight of
Na2CO3=106×19190=10.6 g

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.