Answer:
The volume of the 84 grams of bromine gas at STP is 11.77 L.
Explanation:
Mass of bromine gas = 84 g
Moles of bromine gas = [tex]\frac{84 g}{160 g/mol}=0.525 mol[/tex]
where,
P = Pressure of gas = 1 atm (at STP)
V = Volume of gas = ?
n = number of moles of gas = 0.0525 mol
R = Gas constant = 0.0821 L.atm/mol.K
T = Temperature of gas = 273.15 K (at STP)
Putting values in above equation, we get:
[tex]V=\frac{nRT}{P}[/tex]
[tex]V=\frac{0.525 mol\times 0.0821 atm L/mol K\times 273.15 K}{1 atm}[/tex]
V = 11.77 L
The volume of the 84 grams of bromine gas at STP is 11.77 L.
