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Question

In a mixture of gases, the volume content of a gas is 0.06% at STP. Calculate the number of molecules of the gas in 1 L of the mixture.

A
1.613×1023
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B
6.023×1023
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C
1.61×1027
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D
1.61×1019
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Solution

The correct option is D 1.61×1019
Volume content of gas A in a mixture of A and B =0.06%
1 L of mixture contains= 0.06100×1=6×104 L of A
At STP, 1 mol of gas has volume =22.4 L
number of moles in 6×104 L gas at STP=6×10422.4
number of molecules=number of moles ×6.023×1023
=6×10422.4×6.023×1023
=1.61×1019 molecules

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