Question

The active mass of 64g of HI in two litre flask would be

Answer 1

Active maas will be equal to moles per unit litre volume.

Moles of HI= (given weight) /(molecular weight)

=64/128

=0.5 mole of HI

Active mass=moles/volume(in litre)

=0.5/2

=0.25mol/lit

Answer 2

Answer:

0.25

Explanation:

Active mass = molar concentration = n/v

n=w/hmw

= 64/128

=1/2 = 0.5

n/v =0.5/2 = 0.25

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