Rearrange the following I to IV in the order of increasing masses-I 0-5 moles of O 3II 0-5 g atom of oxygenIII 3-011 - 1023 molecules of O 2 IV 5-6 litre of CO 2 at STPA- II - IV - III - IB- II - I - IV - IIIC- IV - II - III - ID- I - II - III - IV

The correct option is A II lt;IV lt;III lt;I(I) Molar mass of O_3 = 16× 3 g = 48 g Therefore 0.5 moles of O_3=0.5× 48 g = 24 g O_3 (II) 0.5 g atom o ...

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Byju's Answer

Standard XII

Chemistry

Mole Concept

Rearrange the...

Question

Rearrange the following ( $I$ to $I V$ ) in the order of increasing masses:
$(I) 0.5 moles of O_{3}$
$(I I) 0.5 g atom of oxygen$
$(I I I) 3.011 \times 10^{23} molecules of O_{2}$
$(I V) 5.6 litre of C O_{2} at STP$

I I < I V < I I I < I

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I I < I < I V < I I I

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I V < I I < I I I < I

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I < I I < I I I < I V

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Solution

The correct option is A $I I < I V < I I I < I$
$(I)$ $Molar mass of O_{3} = 16 \times 3 g = 48 g$
$Therefore 0.5 moles of O_{3} = 0.5 \times 48 g = 24 g O_{3}$
$(I I)$
$0.5 g atom of oxygen = 0.5 \times 16 g of oxygen = 8 g of oxygen$
$(I I I)$
$3.011 \times 10^{23} Molecules of O_{2} = \frac{3.011 \times 10^{23}}{6.022 \times 10^{23}} \times 32 g of O_{2} = 16 g O_{2}$
$(I V)$
$5.6 litre of C O_{2} at STP = \frac{5.6}{22.4} \times 44 g C O_{2} = 11 g C O_{2}$

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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 litre of CO2 at STP

Rearrange the following ( $I$ to $I V$ ) in the order of increasing masses:
$(I) 0.5 moles of O_{3}$
$(I I) 0.5 g atom of oxygen$
$(I I I) 3.011 \times 10^{23} molecules of O_{2}$
$(I V) 5.6 litre of C O_{2} at STP$