1. Write the definition of molarity

For component 2, $C_2=\frac{\text{moles of component 2}}{\text{volume of solution in litres}}$

If the solution contains $n_1$ moles of component 1 and $n_2$ moles of component 2, then $C_2=\frac{n_2}{V(\text{in L})}$

2. Express solution volume using density

Total mass of solution is $m=n_1M_1+n_2M_2$ (in grams)

Given density $d$ in g mL$^{-1}$, $V(\text{in mL})=\frac{m}{d}=\frac{n_1M_1+n_2M_2}{d}$

So volume in litres is $V(\text{in L})=\frac{n_1M_1+n_2M_2}{1000d}$

Hence, $C_2=\frac{n_2}{\dfrac{n_1M_1+n_2M_2}{1000d}}=\frac{1000d\,n_2}{n_1M_1+n_2M_2}$

3. Use mole fraction of component 2

By definition, $x_2=\frac{n_2}{n_1+n_2}$

Let total moles be $n=n_1+n_2$ Then, $n_2=x_2n, \qquad n_1=(1-x_2)n$

Substitute into the expression for $C_2$: $C_2=\frac{1000d(x_2n)}{(1-x_2)nM_1+x_2nM_2}$

Cancel $n$: $C_2=\frac{1000d x_2}{(1-x_2)M_1+x_2M_2}$

4. Simplify the denominator

$(1-x_2)M_1+x_2M_2=M_1-x_2M_1+x_2M_2$ $=M_1+x_2(M_2-M_1)$

Therefore, $C_2=\frac{1000d x_2}{M_1+x_2(M_2-M_1)}$

5. Match with options

This matches Option B: $\boxed{C_2=\frac{1000d x_2}{M_1+x_2(M_2-M_1)}}$