Question

6. (i) What would be the volume occupied by 32 gm of methane gas at S.T.P? (Atomic mass
of C=12, H=1)
(ii) Convert 590
F to 0
C

Answer 1

Answer:

(i) = 44.8 litre , (ii) = 310°C

Answer 2

1.

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

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

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

2.

=> °F = 1.8 × °C + 32

=> 590 = 1.8×°C +32

=> °C = (590-32)/1.8

=> 590 °F = 310 °C