K4[Fe(CN)6] is supposed to be 40% dissiciated when 1M solution prepared . its boiling point is equal to another 20% mass by volume of non-electrolytic solution A . Considering molality=molarity. The molecular weight of A is

1) 77

2) 67

3) 57

4) 47

K₄[Fe(CN)₆] on dissociation produces 4 K⁺ ions and 1 [Fe(CN)₆]⁴⁺ ions. 
The equation for dissociation is as follows:
K₄[Fe(CN)₆]  → 4 K⁺   + ​[Fe(CN)₆]⁴⁺ 
As we known, $\Delta T_f =i K_f m$
Where symbols have their usual meainings. 

Case I. 
$$\begin{gathered} \underset{\mathrm{C}(1-\mathrm{\alpha })}{\underset{\mathrm{C}}{{\mathrm{K}}_4\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6]}} \to \underset{4\mathrm{C\alpha }}{\underset{0}{4{\mathrm{K}}^{+}}} + \underset{\mathrm{C\alpha }}{\underset{0}{\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6{]}^{4+}}} \\ \mathrm{Hence} \mathrm{number} \mathrm{of} \mathrm{ions} \mathrm{after} \mathrm{dissociation} = \mathrm{C}\left(1-\mathrm{\alpha }\right) +4 \mathrm{C\alpha } + \mathrm{C\alpha } \\ = \mathrm{C} + 4 \mathrm{C\alpha } \\ \mathrm{Number} \mathrm{of} \mathrm{ions} \mathrm{before} \mathrm{dissocaition} = \mathrm{C} \\ \mathrm{i} = \frac{\mathrm{Number} \mathrm{of} \mathrm{ions} \mathrm{after} \mathrm{dissociaion}}{\mathrm{Number} \mathrm{of} \mathrm{ions} \mathrm{before} \mathrm{dissociation}} \\ \mathrm{i} =\frac{\mathrm{C} + 4\mathrm{C\alpha }}{\mathrm{C}} \\ = 1 +4 \mathrm{\alpha } \\ \mathrm{As} \mathrm{\alpha } = 0.40 \\ \mathrm{Therefore} \mathrm{i} = 2.6 \end{gathered}$$

As molality is equal to molarity for this particular case therefore m = 1 M. 
This implies, $\Delta T_f =2.60 K_f 1 M$ .......(i)

Case II. 
For second case, i = 1 as the given solute is non electrolyte. 
The concentration of given solute is 20 % mass by volume, which means that 20 g of solute is dissolved in 100 mL of solution. 
Hence we can write 200 g of solute is present in 1000 mL of solution. 
Molarity of second solution = 200/ M.M 
Where M.M is the molar mass of solute II. 

Since the solvent is same for both the cases therefore K_f in both the cases should be same. 
Moreover, the boiling point of both the solution is same therefore their $\Delta T_f$ should also be equal. 
Hence, we get ​$\Delta T_f =K_f \times \frac{200}{M.M}$......(ii)

On equating both the equations we get, 
$$\begin{gathered} \frac{200}{M.M} =2.60 \\ M.M = \frac{200}{2.60} \\ M.M= 76.92 ~ 77 \end{gathered}$$

Hence, the correct option is 1.