In the given species, decreasing order of the weight of oxygen is:

i) $4.0$ g of $N a O H$
ii) $4.8$ g of $S O_{2}$
iii) $4.0$ g of $C O_{2}$
iv) $2.8$ g of $C a O$

iii > ii > i > iv

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i > ii > iv > iii

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ii > iii > iv > i

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ii > iii > i > iv

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Solution

The correct option is B iii > ii > i > iv
$4.0$ g $N a O H$ corresponds $\frac{4.0}{40} = 0.1$ mole of $N a O H$ .
This is also equal to the number of moles of oxygen atoms.
It weighs $0.1 \times 16 = 1.6 g$ .

$4.8$ g $S O_{2}$ corresponds $\frac{4.8}{64} = 0.075$ mole of $S O_{2}$ .
This is also equal to the number of moles of oxygen molecules $(O_{2})$ .
It weighs $0.075 \times 32 = 2.4 g$ .

$4.0$ g $C O_{2}$ corresponds $\frac{4.0}{44} = 0.09$ mole of $C O_{2}$ .
This is also equal to the number of moles of oxygen molecules $(O_{2})$ .
It weighs $0.09 \times 32 = 2.9 g$ .

$2.8$ g CaO corresponds $\frac{2.8}{56} = 0.05$ mole of $C a O .$
This is also equal to the number of moles of oxygen atoms.
They weigh $0.05 \times 16 = 0.8 g$ .

Hence, the decreasing order of weight of oxygen is iii > ii > i > iv

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In the species

i) 4.0g of $N a O H$ ii) 4.8g of $S O_{2}$

iii) 4.0 g of $C O_{2}$ iv) 2.8 g of $C a O$ ,

decreasing order of the weight of oxygen is

In the species

i) 4.0g of $N a O H$ ii) 4.8g of $S O_{2}$

iii) 4.0 g of $C O_{2}$ iv) 2.8 g of $C a O$ ,

decreasing order of the weight of oxygen is

In the species
i) 4.0g of $N a O H$
ii) 4.8g of $S O_{2}$
iii) 4.0 g of $C O_{2}$
iv) 2.8 g of $C a O$
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Arrange them in decreasing order of the weight of oxygen

i) 4.0g of $N a O H$ ii) 4.8g of $S O_{2}$

iii) 4.0 g of $C O_{2}$ iv)2.8 g of $C a O$

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i) 4.0g of $N a O H$ ii) 4.8g of $S O_{2}$

iii) 4.0 g of $C O_{2}$ iv) 2.8 g of $C a O$ ,

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