Answer:
A) volume of methane = 2.2 l
B) volume of Cl₂ = 2.2 l
Step-by-step explanation:
A) Given data
Number of molecules of methane = 6×10²²
Solution
first we find out the number of moles of methane
number of moles = Number of molecules / Avogadro number
number of moles = 6×10²² / 6.02×10²³
number of moles = 0.1 mol
Now we find out the mass of methane
mass = moles × molar mass
mass of methane = 0.1 mol × 16.04 g/mol
mass of methane = 1.6 g
now we convert it into volume
volume = mass / density
density of methane = 0.72 g/l⁻
volume of methane = 1.6 g / 0.72 g.l⁻
volume of methane = 2.2 l
B) 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
Now we find out the mass of chlorine
mass = moles × molar mass
Molar mass of Cl₂ = 70.9 g/mol
mass of Cl₂ = 0.1 mol × 70.9 g/mol
mass of Cl₂ = 7.1 g
now we convert it into volume
volume = mass / density
density of Cl₂ = 3.2 g/l
volume of Cl₂ = 7.1 g / 3.2 g.l⁻
volume of Cl₂ = 2.2 l