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 $C a O$

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 A
iii, ii, i & iv

i) 4.0g of $N a O H$ contains:

Molecular weight of NaOH = 23 + 16 + 1 = 40 g

40 g of NaOH $\to$ 16 g of Oxygen

4g of NaOH $\to$ 1.6g of Oxygen

ii) 4.8g of $S O_{2}$ contains:

Molecular weight of $S O_{2}$ = 32 + 32 = 64 g

64 g of $S O_{2}$ $\to$ 32 g of Oxygen

4.8g of $S O_{2}$ $\to$ $\frac{4.8 \times 32}{64}$

4.8g of $S O_{2}$ $\to$ 2.4 g of Oxygen

iii) 4.0 g of $C O_{2}$ contains:

Molecular weight of $C O_{2}$ = 12+ 16 + 16 = 44

44 g of $C O_{2}$ $\to$ 32 g of Oxygen

4 g of $C O_{2}$ $\to$ $\frac{4 \times 32}{44}$

4g of $C O_{2}$ $\to$ 2.9 g of Oxygen

iv) 2.8 g of $C a O$ contains:

Molecular weight of $C a O$ = 40+16 = 56

56 g of $C a O$ $\to$ 16 g of Oxygen

2.8 g of $C a O$ $\to$ $\frac{2.8 \times 16}{56}$

2.8 g of $C a O$ $\to$ 0.8g of Oxygen

So, the decreasing order of the weight of oxygen is iii, ii, i & iv

Suggest Corrections

Similar questions

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 $C a O$ ,

decreasing order of the weight of oxygen is

- I

- C N

- N O_{2}

- N H_{2}

- F

+ I

- C H_{3}

- C D_{3}

- C T_{3}

- C H D_{2}

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