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

Active mass means molar concentration of the substance it is expressed by  mol.${\mathrm{L}}^{-1}.$ The formula of active mass is

$\mathrm{Active} \mathrm{mass}=\frac{\mathrm{no}.\mathrm{of} \mathrm{moles}}{\mathrm{Volume} \mathrm{in} \mathrm{L}}=\frac{\mathrm{n}}{\mathrm{V}}$

Given information : 

Volume of solution , V =2 L

Mass of HI , m =64g  

Gram molar mass of HI , GMW=128g/mol

No. of moles of HI,n=$\frac{\mathrm{Weight} \mathrm{of} \mathrm{HI}}{\mathrm{GMW} \mathrm{of} \mathrm{HI}}=\frac{64}{128}=0.5 mol$

$\mathrm{Active} \mathrm{massof} \mathrm{HI}=\frac{\mathrm{No}. \mathrm{of} \mathrm{moles} \left(\mathrm{n}\right)}{\mathrm{Volume} \left(\mathrm{V}\right)\mathrm{in} \mathrm{L}}$

$=\frac{0.5}{2}=0.25 \mathrm{moles}/\mathrm{lit}$

Hence, the correct answer is option “D”.