Water is added to the solution M such that the mole fraction of water in the solution becomes `0.9`. The boiling point of this solution is

Options

A. `380.4 K`
B. `376.2 K`
C. `375.5 K`
D. `354.7 K`

Explanation

To solve the problem, we need to find the boiling point of a solution after water is added, resulting in a mole fraction of water of 0.9. Here’s the step-by-step solution:

### Step 1: Determine the mole fraction of the solute
The mole fraction of water (solvent) is given as 0.9. The mole fraction of the solute can be calculated as:

\[ \text{Mole fraction of solute} = 1 - \text{Mole fraction of solvent} = 1 - 0.9 = 0.1 \]

**Hint:** Remember that the sum of the mole fractions of the solute and solvent must equal 1.

### Step 2: Set up the relationship between moles of solute and solvent
The mole fraction of the solvent can be expressed as:

\[ \text{Mole fraction of solvent} = \frac{\text{moles of solvent}}{\text{moles of solute} + \text{moles of solvent}} = 0.9 \]

And for the solute:

\[ \text{Mole fraction of solute} = \frac{\text{moles of solute}}{\text{moles of solute} + \text{moles of solvent}} = 0.1 \]

**Hint:** Use the definitions of mole fractions to relate the moles of solute and solvent.

### Step 3: Divide the equations for moles of solute and solvent
Dividing the equation for the mole fraction of solute by that of solvent gives:

\[ \frac{\text{moles of solute}}{\text{moles of solvent}} = \frac{0.1}{0.9} \]

This simplifies to:

\[ \frac{\text{moles of solute}}{\text{moles of solvent}} = \frac{1}{9} \]

**Hint:** This ratio indicates how many moles of solute are present relative to the moles of solvent.

### Step 4: Express moles of solute in terms of mass and molar mass
Let the mass of the solute be \( m \) grams and the molar mass of the solute be \( M \) grams/mol. The moles of solute can be expressed as:

\[ \text{moles of solute} = \frac{m}{M} \]

For the solvent (water), the moles can be expressed as:

\[ \text{moles of solvent} = \frac{w}{18} \quad \text{(where \( w \) is the mass of water in grams)} \]

**Hint:** Remember that the molar mass of water is 18 g/mol.

### Step 5: Substitute into the ratio equation
Substituting into the ratio we found:

\[ \frac{m/M}{w/18} = \frac{1}{9} \]

Cross-multiplying gives:

\[ 9m = \frac{Mw}{18} \]

**Hint:** This equation relates the mass of solute to the mass of solvent.

### Step 6: Calculate the boiling point elevation
The boiling point elevation (\( \Delta T_b \)) can be calculated using the formula:

\[ \Delta T_b = K_b \cdot m \]

Where \( K_b \) for water is approximately 0.52 °C kg/mol. The molality \( m \) is defined as:

\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{m/M}{w/1000} \]

Substituting the values we have:

\[ \Delta T_b = 0.52 \cdot \left(\frac{m/M}{w/1000}\right) \]

**Hint:** Ensure you convert grams to kilograms when calculating molality.

### Step 7: Calculate the final boiling point
The boiling point of pure water is 373 K. Thus, the boiling point of the solution is:

\[ T_b = T_{b0} + \Delta T_b = 373 + \Delta T_b \]

Substituting the calculated values will yield the final boiling point.

### Final Calculation
Using the values derived, we can calculate:

\[ \Delta T_b = 0.52 \cdot \left(\frac{0.1}{0.9} \cdot \frac{1000}{18}\right) \]

Calculating this gives:

\[ \Delta T_b = 3.2 \text{ K} \]

Thus, the boiling point of the solution is:

\[ T_b = 373 + 3.2 = 376.2 \text{ K} \]

### Final Answer
The boiling point of the solution is **376.2 K**.