10 grams of limestone on calcination liberates 1.68 lit of CO2 at STP. Percentage purity of limestone is
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answer is 2.
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Detailed Solution
CaCO3S → CaOs + CO2g100 g 22.4 lit10 g 2.24 lit 10 grams of 100% pure limestone liberates 2.24 lit CO2 of at STP10 grams of “X%” pure limestone liberates 1.68 lit of CO2 at STPPercentage purity (X) = 1.682.24 × 100 = 75 %