0.5 g of fuming H 2 S O 4 (Oleum) is diluted with water. This solution is completely neutralized by 26.7 mL of 0.4 N NaOH. The percentage of free S O 3 in the sample is

Question

0.5 g of fuming H 2 S O 4 (Oleum) is diluted with water. This solution is completely neutralized by 26.7 mL of 0.4 N NaOH. The percentage of free S O 3 in the sample is, 30.6%, 40.6%, 20.6%, 50.6%

MARKS App · Accepted Answer

2 N a O H + H 2 S O 4 → N a 2 S O 4 + 2 H 2 O 2 N a O H + S O 3 → N a 2 S O 4 + 2 H 2 O M e q of H 2 S O 4 + M e q of S O 3 = M e q of NaOH ( 0.5 - a ) 49 × 1000 + a 80 / 2 × 1000 = 26.7 × 0.4 So a = 0.103 So % o f S O 3 = 0.103 0.5 × 100 = 20.6 % .

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