Question

How many molecules of ethane are present in 64.28 liters of ethane gas (C2H6) at STP

Answer 1

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 2

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.