Answer: Thus 81 grams of Au is 9210 ml
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
[tex]\text{moles of Au}=\frac{81g}{196.96g/mol}=0.411moles[/tex]
1 mole of Au occupies = 22.4 L
Thus 0.411 moles of Au occupy = [tex]\frac{22.4}{1}\times 0.411=9.21L=9210ml[/tex]
81 grams of Au is 9210 ml
