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What mass of 95% pure CaCO3 will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction ?CaCO3(s)+ 2HCl(aq) -» CaCl2(aq) + CO2(s) + H2O(l) [Calculate upto second place of decimal point]a)1.32 gb)3.65 gc)9.50 gd)1.25 gCorrect answer is option 'A'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about What mass of 95% pure CaCO3 will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction ?CaCO3(s)+ 2HCl(aq) -» CaCl2(aq) + CO2(s) + H2O(l) [Calculate upto second place of decimal point]a)1.32 gb)3.65 gc)9.50 gd)1.25 gCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What mass of 95% pure CaCO3 will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction ?CaCO3(s)+ 2HCl(aq) -» CaCl2(aq) + CO2(s) + H2O(l) [Calculate upto second place of decimal point]a)1.32 gb)3.65 gc)9.50 gd)1.25 gCorrect answer is option 'A'. Can you explain this answer?.

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