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  3. Chemistry
  4. The active mass of 64 g of HI in a two litre flask would be

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Q. The active mass of $64 \,g$ of HI in a two litre flask would be

Equilibrium

Solution:

Active mass is taken as molar concentration i.e number of moles per litre
$\therefore $ Active mass of $HI =\frac{64}{128} \times \frac{1}{2}=0.25$
(Molar mass of $HI =128$ )