Answer: The volume of 0.684 mol of carbon dioxide at s.t.p. is 15.3 L
Explanation:
According to ideal gas equation:
[tex]PV=nRT[/tex]
P = pressure of gas = 1 atm (at STP)
V = Volume of gas = ?
n = number of moles = 0.684
R = gas constant = [tex]0.0821Latm/Kmol[/tex]
T =temperature =[tex]273K[/tex] (at STP)
[tex]V=\frac{nRT}{P}[/tex]
[tex]V=\frac{0.684\times 0.0821L atm/K mol\times 273K}{1atm}[/tex]
[tex]V=15.3L[/tex]
Thus the volume of 0.684 mol of carbon dioxide at s.t.p. is 15.3 L
