The active mass of 64 gm of HI in a two litre flask would be: (A) 2 (B) 1 (C) 5 (D) 0.25

The expression for the active mass is given by

Active mass = (number of moles)/ (volume in litres) = {(mass/molar mass)}/(volume in litres)

We know,

H = 1 g/mol

I = 127 g/mol

On substituting the values, we get,

Active mass = {(64/128)}/2

Active mass = (0.5)/2

We get,

Active mass = 0.25

Hence, the correct option is (D)

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