Question

The active mass of 64 g of $HI$ in a $2-L$ flask would be :

• A. $0.8$
• B. $0.1$
• C. $0.5$
• D. $0.25$

Answer 1

Answer: (D)

$0.25$

Active mass $=$ Concentration $=mol{L}^{-1}$

Molar weight of HI= 128, V= 2 L, m=64 g.

$\left[HI\right]=\frac{64}{128\times 2}=0.25M$