Calculate the number of molecules of oxygen gas that occupies a volume of 224 ml at 273 k and 3 ATM pressure
Answers
Given conditions ⇒
Volume of the Oxygen gas = 224 mL.
Pressure = 3 atm.
Temperature is 273 K.
These conditions are not measured at S.T.P., so we need to convert it into the S.T.P. (Pressure is not measured in S.T.P.)
At S.T.P. Pressure = 1 atm.
Temperature = 273 K.
Volume is to be Measured.
∴ P₁V₁/T₁ = P₂V₂/T₂
⇒ 3 × 224/273 = 1 × V₂/273
⇒ V₂ = 224 × 3
∴ V₂ = 672 mL.
Now, 22400 mL of the Oxygen gas at S.T.P. have the mass of 32 g.
∴ 1 ml of the Oxygen gas at S.T.P. have the mass of 32/22400 g.
∴ 672 ml of the Oxygen gas at S.T.P. have the mass of 32/22400 × 672 g. = 0.96 g.
Now, Using the Formula,
No. of molecules = Mass/Molar Mass × Avogadro's number.
= 0.96/32 × 6.022 × 10²³
= 1.81 × 10²² molecules.
Hence, the number of the molecules of the oxygen gas is 1.81 × 10²².
Hope it helps.