Arrange the following in increasing order of number of molecules
I. 0.5 mol of $H_{2}$ II. 4.0 g of $H_{2}$
III. 18 g of $H_{2} O$ IV. 2.2 g of $C O_{2}$

I I > I I I > I > I V

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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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I V < I I I < I < I I

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Solution

The correct option is D $I V < I < I I I < I I$
1 mol of $H_{2} = 6.02 \times 10^{23}$ molecules of $H_{2}$
0.5 mol of $H_{2} = 3.01 \times 10^{23}$ molecules of $H_{2}$
2g of $H_{2} = 6.02 \times 10^{23}$ molecules of $H_{2}$
4g of $H_{2} = 12.04 \times 10^{23}$ molecules of $H_{2}$
18g of $H_{2} O = 6.023 \times 10^{23}$ molecules of $H_{2} O$
44g of $C O_{2} = 6.023 \times 10^{23}$ molecules of $C O_{2}$
2.2g of $C O_{2} = \frac{2.2 \times 6.023 \times 10^{23}}{44}$ molecules of $C O_{2}$
= $3.01 \times 10^{22}$ molecules of $C O_{2}$
Thus required order will be $I V < I < I I I < I I$

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Similar questions

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 $C a O$

O_{3}

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C O_{2}

I

I V

(I) 0.5 moles of O_{3}

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In the species
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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

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