Active mass of $0.64 g S O_{2}$ In 10 lit vessel is:

1 \times 10^{- 2} M

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1 \times 10^{- 3} M

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1 \times 10^{- 1} M

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0.64 g

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Solution

The correct option is B $1 \times 10^{- 3} M$
$A c t i v e M a s s$ = $\frac{G i v e n W e i g h t}{M o l e c u l a r M a s s \times V o l u m e}$

$A c t i v e M a s s$ = $\frac{0.64}{64 \times 10}$ = $10^{- 3} M$

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