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Question
What is the volume occupied by 48g of oxygen gas at STP?
Solution
Volume occupied by 48g of oxygen:
As 32g of oxygen at STP occupies volume of = 22.4 L
48 g of oxygen at STP occupies volume of = 22.4 x 48/32 = 33.6 L
Hence, 48g of sulphur dioxide will occupy a volume of 33.6L
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