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Academic session 2026-27, Chapter 1 in the current curriculum.
Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter Some Basic Concepts of Chemistry. This page offers a step-by-step solution to the specific question from Exercise 1, Question 35:
Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction,
CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction,
CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?
0.75 M of HCl ≡ 0.75 mol of HCl X molecular weight of HCl dissolved in 1000 ml of water
Or
[(0.75 mol) × (36.5 g mol–1)] HCl is present in 1 L of water
≡ 27.375 g of HCl is present in 1 L of water
Thus, 1000 mL of solution contains Hcl = 27.375g
Or
1 ml of solutions contains Hcl = 27.375/1000 * 1
And 25 ml of solutions contains Hcl = 27.375/1000 * 25 = 0.6844 g.
From the given chemical equation,
CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
2 mol of HCl (2 × 36.5 = 71 g) react with 1 mol of CaCO3 (100 g).
Amount of CaCO3 that will react with 0.6844 g
= 0.9639 g
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 35: Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3(....
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Comments
Thanks teacher it helpsvme a lot..
2x36.5 is 73 and not 71
I like this webpage .... And thanks for helping us preparing our answersðŸâÅ
Thank you very much sir/mam for solving my queries