Answer:
A) V = 11 L
B) V = 24.2 L
C) V = 11.2 L
D) V = 11200 L
Explanation:
A) Given data
Number of moles of methane = 0.5
Solution
Now we find out the mass of methane
mass = moles × molar mass
mass of methane = 0.5 mol × 16.04 g/mol
mass of methane = 8 g
now we convert it into volume
volume = mass / density
density of methane = 0.72 g/l⁻
volume of methane = 8 g / 0.72 g.l⁻
volume of methane = 11 L
B) Given data
Number of moles of chlorine = 0.5
Solution
1st we find out the mass of chlorine
mass = moles × molar mass
Molar mass of Cl₂ = 70.9 g/mol
mass of Cl₂ = 0.5 mol × 70.9 g/mol
mass of Cl₂ = 35.5 g
now we convert it into volume
volume = mass / density
density of Cl₂ = 1.468 g/l
volume of Cl₂ = 35.5 g / 1.468 g.l⁻
volume of Cl₂ = 24.2 L
C) Given data
Number of moles of chlorine = 0.5 mol
Solution
Volume of chlorine:
Standard volume of one mole of gas is 22.4 L.
For 0.5 mol:
22.4 × 0.5 = 11.2 L
Another method:
At the standard volume gas occupy the standard pressure and temperature:
PV = nRT
V = nRT / P
V = 0.5 mol × 0.0821 L. atm. mol⁻¹. k⁻¹× 273 K / 1 atm
V = 11.2 L.atm / 1 atm
V = 11.2 L
D) Given data
Number of moles of chlorine = 0.5 kmol
Number of moles of chlorine = 0.5 kmol × 1000
Number of moles of chlorine = 500 mol
Solution
Volume of chlorine:
Standard volume of one mole of gas is 22.4 L.
For 500 mol:
22.4 × 500 = 11200 L
Another method:
At the standard volume gas occupy the standard pressure and temperature:
PV = nRT
V = nRT / P
V = 500 mol × 0.0821 L. atm. mol⁻¹. k⁻¹× 273 K / 1 atm
V = 11200 L.atm / 1 atm
V = 11200 L
