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Question

21 g of nitrogen gas is mixed with 1.2 moles of SO2 gas and 1.21×1023 molecules of O2 gas. Find the volume of the gaseous mixture (assuming no reaction) at STP.

A
22.4 L
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B
15.67 L
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C
48.16 L
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D
67.56 L
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Solution

The correct option is C 48.16 L
Moles of nitrogen =2128=0.75 mole
1 mole of O2 gas contains 6.023×1023 molecules
So, 1.21×1023 are present in 0.2 moles.
Total moles of gases =0.75+1.2+0.2=2.15 moles.
1 mole occupies 22.4 L of volume
2.15 mole occupies =22.4×2.15=48.16 L

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