A solution of two components containing $$n_1$$ moles of the 1st component and $$n_2$$ moles of the 2nd component is prepared. $$M_1$$ and $$M_2$$ are the molecular weights of component 1 and 2 respectively. If $$d$$ is the density of the solution in $$\text{g mL}^{-1}$$, $$C_2$$ is the molarity and $$x_2$$ is the mole fraction of the 2nd component, then $$C_2$$ can be expressed as:

We have a binary solution that contains $$n_1$$ moles of component 1 and $$n_2$$ moles of component 2. The respective molecular weights are $$M_1$$ and $$M_2$$ (in $$\text{g mol}^{-1}$$). The density of the entire solution is given as $$d$$ in $$\text{g mL}^{-1}$$. The problem asks us to obtain an expression for the molarity $$C_2$$ of component 2 in terms of the given quantities and the mole fraction $$x_2$$ of the second component.

First we recall the definition of molarity.

Formula stated:  $$\text{Molarity}\;(C) \;=\;\dfrac{\text{moles of solute}}{\text{volume of solution in litres}}$$.

Here the “solute” is component 2, so

$$C_2 \;=\;\dfrac{n_2}{V_{\text{solution (in L)}}}.$$

To evaluate the volume of the solution, we use its density. Density is defined as

Formula stated:  $$d \;=\;\dfrac{\text{mass of solution}}{\text{volume of solution in mL}}.$$

So the total volume in millilitres is

$$V_{\text{(in mL)}} \;=\;\dfrac{\text{mass of solution}}{d}.$$

Because molarity requires the volume in litres, we convert millilitres to litres by dividing by $$1000$$:

$$V_{\text{(in L)}} \;=\;\dfrac{\text{mass of solution}}{1000\,d}.$$

Substituting this volume into the molarity expression we obtain

$$C_2 \;=\;\dfrac{n_2}{\dfrac{\text{mass of solution}}{1000\,d}} \;=\;\dfrac{1000\,d\,n_2}{\text{mass of solution}}.$$

Now we must write the mass of the solution entirely in terms of $$n_2$$ and the mole fraction $$x_2$$, so that $$n_2$$ can cancel later.

Total mass of the solution arises from both components:

$$\text{mass of solution} \;=\; n_1M_1 \;+\; n_2M_2.$$

To eliminate $$n_1$$ we employ the definition of the mole fraction.

Formula stated:  $$x_2 \;=\;\dfrac{n_2}{n_1+n_2}.$$

Rearranging for $$n_1$$ gives

$$n_1 \;=\;\dfrac{n_2(1-x_2)}{x_2}.$$

Substituting this value of $$n_1$$ in the mass expression:

$$\text{mass} \;=\;\bigg[\dfrac{n_2(1-x_2)}{x_2}\bigg]M_1 \;+\; n_2M_2.$$

We now factor out $$n_2$$ for clarity:

$$\text{mass} \;=\; n_2\Bigg[\dfrac{(1-x_2)M_1}{x_2} \;+\; M_2\Bigg].$$

Placing the two terms over the common denominator $$x_2$$ we get

$$\text{mass} \;=\; n_2\Bigg[\dfrac{M_1(1-x_2) \;+\; M_2x_2}{x_2}\Bigg] \;=\; n_2\Bigg[\dfrac{M_1 \;+\; x_2(M_2-M_1)}{x_2}\Bigg].$$

Thus

$$\text{mass of solution} \;=\;\dfrac{n_2\bigl[M_1 + x_2(M_2-M_1)\bigr]}{x_2}.$$

We now substitute this mass back into the molarity relation obtained earlier:

$$C_2 \;=\;\dfrac{1000\,d\,n_2}{\dfrac{n_2\bigl[M_1 + x_2(M_2-M_1)\bigr]}{x_2}} \;=\;1000\,d\,n_2 \times \dfrac{x_2}{n_2\bigl[M_1 + x_2(M_2-M_1)\bigr]}.$$

Notice that $$n_2$$ in the numerator and denominator cancels out, yielding

$$C_2 \;=\;\dfrac{1000\,d\,x_2}{M_1 + x_2(M_2 - M_1)}.$$

This is precisely the form given in Option C.

Hence, the correct answer is Option C.