$8 \ g$ of $H_2$ and $256 \ g$ of $HI$ are in a $4 \ L$ flask. Calculate their active mass. $(H = 1 \ g \ mol^{-1}, I = 127 \ g \ mol^{-1})$

(A) Active mass is defined as molar concentration,which is given by $\text{Active mass} = \frac{\text{Number of moles}}{\text{Volume in } L}$.
For $H_2$: Molar mass $= 2 \times 1 = 2 \ g \ mol^{-1}$. Moles $= \frac{8 \ g}{2 \ g \ mol^{-1}} = 4 \ mol$. Active mass $= \frac{4 \ mol}{4 \ L} = 1 \ mol \ L^{-1}$.
For $HI$: Molar mass $= 1 + 127 = 128 \ g \ mol^{-1}$. Moles $= \frac{256 \ g}{128 \ g \ mol^{-1}} = 2 \ mol$. Active mass $= \frac{2 \ mol}{4 \ L} = 0.5 \ mol \ L^{-1}$.