Step-by-step Derivation:

1. Analyze the given information:

   • Mole fraction of solute, $x_{solute} = 0.1$.
   • The molarity (M) of the solution is equal to its molality (m).
   • Density of the solution, $d = 2.0 \text{ g cm}^{-3} = 2.0 \text{ g/mL}$.
   • We need to find the ratio of molecular weights: ${{M{W_{solute}}} \over {M{W_{solvent}}}}$.

2. Relate mole fraction to moles:

   • The sum of mole fractions of all components in a solution is 1. Therefore, the mole fraction of the solvent is: $x_{solvent} = 1 - x_{solute} = 1 - 0.1 = 0.9$
   • Let's assume we have a sample of the solution containing $n_{solute}$ moles of solute and $n_{solvent}$ moles of solvent. The ratio of their moles is equal to the ratio of their mole fractions: ${{n_{solute}} \over {n_{solvent}}} = {{x_{solute}} \over {x_{solvent}}} = {{0.1} \over {0.9}} = {1 \over 9}$
   • For simplicity in calculations, let's assume we have 1 mole of solute ($n_{solute} = 1$ mol). Then, the number of moles of solvent would be $n_{solvent} = 9$ mol.

3. Express Molarity and Molality in terms of molecular weights:

   • Let $M_1$ be the molecular weight of the solute ($MW_{solute}$) and $M_2$ be the molecular weight of the solvent ($MW_{solvent}$).
   • Mass of solute ($w_1$): $w_1 = n_{solute} \times M_1 = 1 \times M_1 = M_1$ g.
   • Mass of solvent ($w_2$): $w_2 = n_{solvent} \times M_2 = 9 \times M_2 = 9M_2$ g.
   • Total mass of the solution ($W_{sol}$): $W_{sol} = w_1 + w_2 = (M_1 + 9M_2)$ g.
   • Volume of the solution ($V_{sol}$): Using the density $d$, $V_{sol} = {{W_{sol}} \over d} = {{M_1 + 9M_2} \over {2.0}} \text{ mL}$

4. Calculate Molarity (M) and Molality (m):

   • Molarity (M): Moles of solute per liter of solution. $M = {{n_{solute}} \over {V_{sol} \text{ (in L)}}} = {{1} \over {({{M_1 + 9M_2} \over {2.0}}) / 1000}} = {{2000} \over {M_1 + 9M_2}}$
   • Molality (m): Moles of solute per kilogram of solvent. $m = {{n_{solute}} \over {w_2 \text{ (in kg)}}} = {{1} \over {(9M_2) / 1000}} = {{1000} \over {9M_2}}$

5. Apply the given condition (Molarity = Molality):

   • We are given that $M = m$. ${{2000} \over {M_1 + 9M_2}} = {{1000} \over {9M_2}}$

6. Solve for the required ratio:

   • Divide both sides by 1000: ${{2} \over {M_1 + 9M_2}} = {1 \over {9M_2}}$
   • Cross-multiply to solve for the ratio $M_1/M_2$: $2 \times (9M_2) = 1 \times (M_1 + 9M_2)$ $18M_2 = M_1 + 9M_2$
   • Rearrange the terms: $18M_2 - 9M_2 = M_1$ $9M_2 = M_1$
   • Therefore, the ratio of the molecular weights is: ${{M_1} \over {M_2}} = {{M{W_{solute}}} \over {M{W_{solvent}}}} = 9$

Final Answer

The ratio of the molecular weights of the solute and solvent is 9.