A) what is the volume of 0.5 mol methane in normal state?
b) what is the volume of 0.5 mol chlorine gas in normal state?
c) what is the volume of 0.5 mol chlorine gas in standard state?
d) what is the volume of 0.5 kmol chlorine gas in standard state?

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