Mole fraction of a non-electrolyte in aqueous solution is 0.07. If  $K_f\left({\mathrm{H}}_2\mathrm{O}\right)$ is ${1.86}^{\circ }{\mathrm{mol}}^{-1}\mathrm{kg}$ depression in f. p.$\mathrm{\Delta }{T}_f$ is:

Given,

Mole fraction of a non electrolyte in aqueous solution = 0.07

${\mathrm{K}}_{\mathrm{f}}=1.86$

Depression in freezing point = ?

Mole fraction of solvent = 1- mole fraction of non electrolyte

$\Rightarrow {\mathrm{X}}_1=1-{\mathrm{X}}_2=1-0.07=0.93$

Depression in freezing  point,.

${\mathrm{\Delta T}}_{\mathrm{f}}=\mathrm{i}\times \mathrm{m}\times {\mathrm{K}}_{\mathrm{f}}$

Since the solute is non electrolyte, i=1

$\text{ Molarity, }m=\frac{\text{ No. Of moles of solute }}{\text{ Mass of solvent in }\mathrm{Kg}}$

$$\begin{gathered} \mathrm{\Delta }{T}_f=\frac{1.86\times 0.07\times 1000}{0.93\times 18} \\ =7.78\mathrm{K} \end{gathered}$$

Therefore, ${\mathrm{\Delta T}}_{\mathrm{f}}=1\times 4.18\times 1.86={7.78}^{\circ }\mathrm{C}$

Hence, correct option  is D.