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 solvent given the mole fraction of the solute, the density of the solution, and the condition that molarity is equal to molality.

### Step-by-Step Solution:

1. **Understanding Given Data:**
- Mole fraction of solute, \( X_{solute} = 0.1 \)
- Density of the solution, \( \rho = 2.0 \, \text{g/cm}^3 \)
- Temperature, \( T = 298 \, \text{K} \)

2. **Assuming a Volume of Solution:**
- Let’s assume we have 1 liter (1000 mL) of the solution.
- The mass of the solution can be calculated using its density:
\[ \text{Mass of solution} = \text{Density} \times \text{Volume} = 2.0 \, \text{g/cm}^3 \times 1000 \, \text{cm}^3 = 2000 \, \text{g} \]

3. **Calculating Mass of Solute and Solvent:**
- Let the mass of the solute be \( m_{solute} = x \, \text{g} \).
- Then, the mass of the solvent will be:
\[ m_{solvent} = 2000 \, \text{g} - x \, \text{g} \]

4. **Finding Moles of Solute and Solvent:**
- The number of moles of solute is given by:
\[ n_{solute} = \frac{x}{M_{solute}} \quad \text{(where \( M_{solute} \) is the molecular weight of the solute)} \]
- The number of moles of solvent is given by:
\[ n_{solvent} = \frac{2000 - x}{M_{solvent}} \quad \text{(where \( M_{solvent} \) is the molecular weight of the solvent)} \]

5. **Using Mole Fraction:**
- The mole fraction of the solute is given by:
\[ X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}} = 0.1 \]
- Substituting the expressions for moles:
\[ 0.1 = \frac{\frac{x}{M_{solute}}}{\frac{x}{M_{solute}} + \frac{2000 - x}{M_{solvent}}} \]

6. **Cross-Multiplying and Simplifying:**
- Cross-multiplying gives:
\[ 0.1 \left( \frac{x}{M_{solute}} + \frac{2000 - x}{M_{solvent}} \right) = \frac{x}{M_{solute}} \]
- Rearranging leads to:
\[ 0.1 \cdot \frac{2000 - x}{M_{solvent}} = 0.9 \cdot \frac{x}{M_{solute}} \]

7. **Solving for the Ratio of Molecular Weights:**
- Rearranging gives:
\[ 0.1 M_{solute} (2000 - x) = 0.9 x M_{solvent} \]
- Solving for \( \frac{M_{solute}}{M_{solvent}} \):
\[ M_{solute} = 9 M_{solvent} \]
- Therefore, the ratio of the molecular weights is:
\[ \frac{M_{solute}}{M_{solvent}} = 9 \]

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