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The incorrect statement for 14g of CO is

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It occupies 2.24 L at STP

It corresponds to ½ mol of CO

It corresponds to same number of moles as 14g $N_{2}$

It corresponds to 3.01 x $10^{23}$ molecules of CO

answer is A.

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Detailed Solution

14g $C O = 14 / 28 = 0.5$ mol
It has a volume of 11.2L at STP; it has same number of moles as 14g $N_{2}$ (0.5 mol); has $3.01 \times 10^{23}$ molecules

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