What is the degree of freedom for two mis-cible (non-reacting) substances in vapor-liquid equilibrium forming an azeotrope?

Answer: Option C

The given number of component $$=2$$
Given number of phases $$=2$$
We know, $$\text{degree of freedom (number of independent variables) }=C-\varphi +2$$
$$\left(\text{assuming no reaction taking place}\right)$$
$$f=2-2+2=2$$
And since under azeotropic conditions the composition of a component in both the phases remains same one of the independent variable becomes dependent and hence the degree of freedom reduces to 1.