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QNEET-2022-Ch-368: NEET 2022 Chemistry - Class 11 NEET-2022-Ch-368 Solution: What mass of 95% pure \(CaCO_3\) will be required to neutralise …
What mass of 95% pure will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction? [Calculate upto second place of decimal point] 1.25 g 1.32 g 3.65 g 9.50 g
Multiple Choice Options
- A 1.25 g
- B 1.32 g
- C 3.65 g
- D 9.50 g
Step-by-Step Verified Solution
Let m gram mass of CaCO 3 is required
Pure CaCO 3 in m gram = 95 100 _ m
Moles of CaCO 3 = 95 100 _ m 100
Moles of HCl required = 2 _ moles of CaCO 3
= 2 _ 95 100 _ m 100
2 _ 95 100 _ m 100 = 50 1000 _ 0.5
m = 1.315 g � 1.32 g