A solution containing `2.5 xx 10^(-3)` kg of a solute dissolved in `75xx 10^(-3)` kg of water boils at 373.535 K . The molar mass of the solute is ______ g `"mol"^(-1)` [nearest boiling point of water = 373.15K)

Explanation

To find the molar mass of the solute, we can use the formula for the boiling point elevation. The boiling point elevation (\( \Delta T_b \)) is given by the equation:

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

Where:
- \( \Delta T_b \) is the boiling point elevation,
- \( K_b \) is the ebullioscopic constant of the solvent (water in this case),
- \( m \) is the molality of the solution.

### Step 1: Calculate the boiling point elevation

The boiling point of pure water is 373.15 K, and the boiling point of the solution is 373.535 K. Therefore, the boiling point elevation is:

\[ \Delta T_b = 373.535 \, K - 373.15 \, K = 0.385 \, K \]

### Step 2: Find the molality of the solution

The molality (\( m \)) is defined as the number of moles of solute per kilogram of solvent. We can rearrange the boiling point elevation equation to solve for molality:

\[ m = \frac{\Delta T_b}{K_b} \]

Given that the ebullioscopic constant \( K_b \) for water is approximately \( 0.512 \, K \cdot kg/mol \):

\[ m = \frac{0.385 \, K}{0.512 \, K \cdot kg/mol} \approx 0.752 \, mol/kg \]

### Step 3: Calculate the number of moles of solute

We know the mass of the solute is \( 2.5 \times 10^{-3} \, kg \) and the mass of the solvent (water) is \( 75 \times 10^{-3} \, kg \).

Using the definition of molality:

\[ m = \frac{n}{mass \, of \, solvent \, (kg)} \]

Where \( n \) is the number of moles of solute. Rearranging gives:

\[ n = m \cdot mass \, of \, solvent \]

Substituting the values:

\[ n = 0.752 \, mol/kg \cdot 0.075 \, kg \approx 0.0564 \, mol \]

### Step 4: Calculate the molar mass of the solute

Molar mass (\( M \)) is defined as the mass of the solute divided by the number of moles of solute:

\[ M = \frac{mass \, of \, solute}{n} \]

Substituting the values:

\[ M = \frac{2.5 \times 10^{-3} \, kg}{0.0564 \, mol} \approx 0.0443 \, kg/mol \]

To convert this to grams per mole:

\[ M \approx 44.3 \, g/mol \]

### Final Answer

The molar mass of the solute is approximately **44.3 g/mol**.

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