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Law of Mass Action

The active ma...

Question

The active mass of $64 g$ of $H I$ in a $2 L$ flask would be

A

2

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B

1

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C

5

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D

0.25

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Solution

The correct option is D 0.25
Molar mass of $H I = 127 + 1 = 128 g$
Moles of $H I = \frac{64}{128} = 0.5$
volume= $2 L$
$[H I] = \frac{M o l e s}{v o l u m e}$
putting the values.
$[H I] = \frac{0.5}{2} = 0.25$

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Law of Mass Action

Standard XII Chemistry

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