Rearrange the following (I to IV) in the order of increasing masses.(I) 0.5 mol of O3 (II) 0.5 g molecule of Nitrogen(III) 3.011 × 1023 molecules of O2 (IV) 11.35 L of CO2 at STP.

Engineering

Chemistry

Mole and Its Calculation

Question

Rearrange the following (I to IV) in the order of increasing masses.

(I) 0.5 mol of O₃

(II) 0.5 g molecule of Nitrogen

(III) 3.011 × 10²³ molecules of O₂

(IV) 11.35 L of CO₂ at STP.

IV < III < II < I

II < III < IV < I

I < II < III < IV

III < II < I < IV

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(I)   0.5 mol of O₃ ⇒ mass = 0.5 × 48 = 24 g
(II)   0.5 g molecule of N₂ ⇒ mass = 0.5 × 28 = 14 g
(III)  3.011 × 10²³ molecules of O₂ ⇒ mass = 0.5 × 32 = 16 g
(IV) 11.35 L of CO₂ at STP ⇒ mass = 0.5 × 44 = 22 g