Verified Answer :

Use the Ideal Gas Law
\(PV = nRT\)
\((1.00atm)(64.28L) = n(.08206L*atm*mol^{-1}*K^{-1})(273.15K)\)
n = 2.8677moles
\(2.8677mol* \frac{6.022*10^{23}molecules}{1mol}\)
\(= 1.727*10^{24}molecules\)
That was using Avogadro's number :)

Answer : The molecule of ethane present in 64.28 L of ethane gas at STP is, \(17.277\times 10^{23}\) molecule.

Solution :

At STP,

22.4 L volume of ethane present in 1 mole of ethane gas

64.28 L volume of ethane present in \(\frac{64.28L}{22.4L}\times 1mole=2.869moles\) of ethane gas

And, as we know that

1 mole of ethane molecule contains \(6.022\times 10^{23}\) molecules of ethane

2.869 moles of ethane molecule contains \(2.869\times 6.022\times 10^{23}=17.277\times 10^{23}\) molecules of ethane

Therefore, the molecule of ethane present in 64.28 L of ethane gas at STP is, \(17.277\times 10^{23}\) molecule.

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