Question

Calculate the volume in dm3 occupied by 32 gram methane at S.T.P.

Answer 1

The volume in dm3 occupied by 32-gram methane at S.T.P. is 44.82 dm³.

Explanation:

Mass of methane, m = 32 gm

Molecular mass of methane, M = 16 g/mol

∴ No. of moles, n = m/M = 32/16 = 2 moles

At S.T.P. conditions, we have

Pressure, P = 1 atm

Temperature, T = 273 K

Let the volume of methane gas be “V”.

Now,

Using the Ideal Gas Law,

PV = nRT

⇒ 1 * V = 2 * 0.0821 * 273 …… [R = ideal gas constant = 0.0821 L atm K⁻¹ mol⁻¹]

⇒ V = 44.82 L

We know, 1 L = 1 dm³

∴ V = 44.82 L = 44.82 dm³

Thus, 32 grams of methane at S.T.P will occupy 44.82 dm³.

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Answer 2

Answer:

Explanation:

∴ No. of moles, n = m/M = 32/16 = 2 moles

At S.T.P. conditions, we have

Pressure, P = 1 atm

Temperature, T = 273 K

Let the volume of methane gas be “V”.

Now,

Using the Ideal Gas Law,

PV = nRT

⇒ 1 * V = 2 * 0.0821 * 273 …… [R = ideal gas constant = 0.0821 L atm K⁻¹ mol⁻¹]

⇒ V = 44.82 L

We know, that 1 L = 1 dm³

∴ V = 44.82 L = 44.82 dm³

Thus, 32 grams of methane at S.T.P will occupy 44.82 dm³.