Correct option is (D) 0.25
The expression for the active mass is,
\(\text{Active mass} = \frac{\text{number of moles}}{\text{volume in litres}} = \frac{\frac{\text{mass}}{\text{molar mass}}}{\text{volume in litres}} \)
Substitute values in the above expression,
\(\text{Active mass} = \frac{\frac{64}{128}}{2} = 0.25\)
