Gram molecular mass of O2 = 2 × 16 = 32 g
1 mole of O2 weighs 32 g and occupies 22.4 lit. vol.
∴ 3.5 g of O2 occupies = = 22.4\over 32 × 3.5
Volume occupied by 3.5 g of O2 gas at 27°C and 740 mm pressure:
| s.t.p. | given values |
|---|---|
| P1 = 760 mm of Hg | P2 = 740 mm of Hg |
| V1 = 2.45 lit | V2 = x lit |
| T1 = 273 K | T2 = 27 + 273 K |
Using the gas equation,
⇒ P_1V_1\over T_1=P_2V_2\over T_2
Substituting the values we get,
⇒ 760 × 2.45\over 273=740× x\over 300
⇒ x = 760 × 2.45 × 300\over 740 × 273
x = 2.76 lit
Hence, the volume occupied by 3.5 g of O2 gas at 27°C and 740 mm pressure is 2.76 lit.
