Answer:
A) volume of Cl₂ = 2.24 L
B) volume of methane = 134.4 L
Step-by-step explanation:
A) Given data
Number of molecules of chlorine = 6×10²²
Solution
first we find out the number of moles of chlorine
number of moles = Number of molecules / Avogadro number
number of moles = 6×10²² / 6.02×10²³
number of moles = 0.1 mol
Volume of chlorine:
Standard volume of one mole of gas is 22.4 L.
For 0.1 mol:
22.4 × 0.1 L = 2.24 L
Another method:
At the standard volume gas occupy the standard pressure and temperature:
PV = nRT
V = nRT / P
V = 0.1 mol. 0.0821 L. atm. mol⁻¹. k⁻¹× 273 K / 1 atm
V = 2.24 L.atm / 1 atm
V = 2.24 L
B) Given data
Number of molecules of methane = 3.6×10²⁴
Solution
first we find out the number of moles of methane
number of moles = Number of molecules / Avogadro number
number of moles = 3.6×10²⁴ / 6.02×10²³
number of moles = 6.0 mol
Now we find out the volume of methane
Standard volume of one mole of gas is 22.4 L.
For 6.0 mol:
22.4 × 6.0 L = 134.4 L
Another method:
At the standard volume gas occupy the standard pressure and temperature:
PV = nRT
V = nRT / P
V = 6 mol. 0.0821 L. atm. mol⁻¹. k⁻¹× 273 K / 1 atm
V = 134.4 L. atm/ 1 atm
V = 134.4 L