The mole fraction of a solute in a solutions is `0.1`. At `298K` molarity of this solution is the same as its molality. Density of this solution at 298 K is `2.0 g cm^(-3)`. The ratio of the molecular weights of the solute and solvent, `(MW_("solute"))/(MW_("solvent"))` is

Explanation

To solve the problem, we need to find the ratio of the molecular weights of the solute and the solvent given the mole fraction of the solute, the molarity, the molality, and the density of the solution.

### Step-by-Step Solution:

1. **Understanding the Given Data**:
- Mole fraction of solute, \( x_1 = 0.1 \)
- Molarity (M) = Molality (m)
- Density of solution, \( \rho = 2.0 \, \text{g/cm}^3 \)

2. **Calculate the Mole Fraction of Solvent**:
\[ x_2 = 1 - x_1 = 1 - 0.1 = 0.9 \]

3. **Define Molarity and Molality**:
- Molarity (M) is defined as:
\[ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]
- Molality (m) is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]

4. **Using Density to Find Volume**:
- The mass of the solution can be calculated using density:
\[ \text{mass of solution} = \text{volume} \times \text{density} \]
- Let’s assume we have 1 liter of solution (1000 cm³):
\[ \text{mass of solution} = 1000 \, \text{cm}^3 \times 2.0 \, \text{g/cm}^3 = 2000 \, \text{g} \]

5. **Calculate the Mass of Solute and Solvent**:
- Let \( M_1 \) be the molecular weight of the solute and \( M_2 \) be the molecular weight of the solvent.
- Mass of solute:
\[ \text{mass of solute} = x_1 \times \text{mass of solution} = 0.1 \times 2000 \, \text{g} = 200 \, \text{g} \]
- Mass of solvent:
\[ \text{mass of solvent} = x_2 \times \text{mass of solution} = 0.9 \times 2000 \, \text{g} = 1800 \, \text{g} \]

6. **Convert Mass of Solvent to Kilograms**:
\[ \text{mass of solvent in kg} = \frac{1800 \, \text{g}}{1000} = 1.8 \, \text{kg} \]

7. **Calculate Moles of Solute**:
\[ \text{moles of solute} = \frac{\text{mass of solute}}{M_1} = \frac{200 \, \text{g}}{M_1} \]

8. **Set Up the Equation for Molarity**:
Since molarity equals molality:
\[ M = m \implies \frac{200/M_1}{1} = \frac{200/M_1}{1.8} \]

9. **Cross-Multiply and Simplify**:
\[ 200 \times 1.8 = 200 \implies 1.8 = \frac{200}{M_1} \implies M_1 = \frac{200}{1.8} \]

10. **Calculate the Ratio of Molecular Weights**:
\[ \frac{M_1}{M_2} = \frac{200/1.8}{M_2} \]
From the mass of solvent:
\[ M_2 = \frac{1800}{\text{moles of solvent}} = \frac{1800}{0.9 \times \frac{200}{M_1}} \implies M_2 = \frac{1800 \times M_1}{180} \implies M_2 = 10M_1 \]
Thus:
\[ \frac{M_1}{M_2} = \frac{1}{10} \implies \frac{M_1}{M_2} = 9:1 \]

### Final Answer:
The ratio of the molecular weights of the solute and solvent is \( \frac{M_{\text{solute}}}{M_{\text{solvent}}} = 9:1 \).