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Mole Concept and Avogadro's Number

4g H 2, 32...

Question

$4$ g $H_{2}$ , $32$ g $O_{2}$ , $14$ g $N_{2}$ and $11$ g ${C O}_{2}$ are taken in a bulb of $500$ ml. Which one of these has maximum active mass?

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Solution

$A c t i v e M a s s = \frac{G i v e n M a s s}{M o l e c u l a r M a s s \times V o l u m e}$

The Higher the Ratio of Given mass and Molecular weight, higher will be the Active Mass

For $H_{2}$ , the ratio is 2
For $O_{2}$ , the ratio is 1
For $N_{2}$ , the ratio is 0.5
For $C O_{2}$ , the ratio is 0.25

So 4g of $H_{2}$ will have the maximum active mass.

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Similar questions

4 g H_{2}, 32 g O_{2}, 14 g N_{2}

11 g C O_{2}

500 m l

$4 g H_{2}, 32 g O_{2}, 14 g N_{2}$ and $11 g C O_{2}$ are taken in a bulb of 500 ml. which one of these has maximum active mass?

4 g

H_{2}

32 g

O_{2}

14 g

N_{2}

11 g

{C O}_{2}

500 m l

22

C O_{2}

3 g

H_{2}

7 g

N_{2}

14

N_{2}

64

O_{2}

8

H e

35

{C l}_{2}

380

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