Respuesta :

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

Answer : The molecule of ethane present in 64.28 L of ethane gas at STP is, [tex]17.277\times 10^{23}[/tex] 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 [tex]\frac{64.28L}{22.4L}\times 1mole=2.869moles[/tex] of ethane gas

And, as we know that

1 mole of ethane molecule contains [tex]6.022\times 10^{23}[/tex] molecules of ethane

2.869 moles of ethane molecule contains [tex]2.869\times 6.022\times 10^{23}=17.277\times 10^{23}[/tex] molecules of ethane

Therefore, the molecule of ethane present in 64.28 L of ethane gas at STP is, [tex]17.277\times 10^{23}[/tex] molecule.

Otras preguntas