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Byju's Answer

Standard XIII

Chemistry

Mole Concept

Rearrange the...

Question

Rearrange the following (I to IV) in the order of increasing masses.
(I) 0.5 moles of $O_{3}$
(II) $0.5$ g-atom of oxygen
(III) $3.011 \times 10^{23}$ molecules of $O_{2}$
(IV) $5.6 L$ of $C O_{2}$ at STP

II < IV < III < I

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II < I < IV < III

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IV < II < III < I

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I < II < III < IV

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Solution

The correct option is A II < IV < III < I
(I) 0.5 moles $O_{3} = 24 g$ of $O_{3}$

(II) $0.5 g$ atom of oxygen = $8 g$

(III) $\frac{3.011 \times 10^{23}}{6.022 \times 10^{23}} \times 32 g O_{2} = 16 g$

(IV) $\frac{5.6}{22.4} \times 44 g C O_{2} = 11 g C O_{2}$

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Rearrange the following (I to IV) in the order of increasing masses. (I) 0.5 moles of O3 (II) 0.5 g-atom of oxygen (III) 3.011×1023 molecules of O2 (IV) 5.6 L of CO2 at STP

The correct option is A II < IV < III < I(I) 0.5 moles O3=24 g of O3 (II) 0.5 g atom of oxygen = 8 g (III) 3.011×10236.022×1023×32 g O2=16 g (IV) 5.622.4×44 g CO2=11 g CO2

Rearrange the following (I to IV) in the order of increasing masses.
(I) 0.5 moles of $O_{3}$
(II) $0.5$ g-atom of oxygen
(III) $3.011 \times 10^{23}$ molecules of $O_{2}$
(IV) $5.6 L$ of $C O_{2}$ at STP

The correct option is A II < IV < III < I
(I) 0.5 moles $O_{3} = 24 g$ of $O_{3}$

(II) $0.5 g$ atom of oxygen = $8 g$

(III) $\frac{3.011 \times 10^{23}}{6.022 \times 10^{23}} \times 32 g O_{2} = 16 g$

(IV) $\frac{5.6}{22.4} \times 44 g C O_{2} = 11 g C O_{2}$