A solution containing 12.5 g of non-electrolyte substance in 185 g of water shows boiling point elevation of 0.80 K. Calculate the molar mass of the substance. (`K_b=0.52 K kg mol^(-1)`)

Options

A. `"53.06 g mol"^(-1)`
B. `"25.3 g mol"^(-1)`
C. `"16.08 g mol"^(-1)`
D. `"43.92 g mol"^(-1)`

Explanation

To find the molar mass of the non-electrolyte substance, we can use the formula for boiling point elevation, which is given by:

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

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

### Step-by-Step Solution:

**Step 1: Identify the given values.**
- Mass of the solute (non-electrolyte substance) = 12.5 g
- Mass of the solvent (water) = 185 g = 0.185 kg (convert grams to kilograms)
- Boiling point elevation (\(\Delta T_b\)) = 0.80 K
- \(K_b\) for water = 0.52 K kg mol\(^{-1}\)

**Step 2: Calculate the molality (m) of the solution.**
Molality is defined as the number of moles of solute per kilogram of solvent.

First, we need to express the number of moles of the solute:
\[ \text{Number of moles of solute} = \frac{\text{mass of solute}}{\text{molar mass of solute}} = \frac{12.5 \, \text{g}}{M} \]
where \(M\) is the molar mass of the solute.

Now, we can calculate the molality:
\[ m = \frac{\text{Number of moles of solute}}{\text{mass of solvent in kg}} = \frac{\frac{12.5}{M}}{0.185} \]

**Step 3: Substitute the molality into the boiling point elevation equation.**
Substituting \(m\) into the boiling point elevation equation:
\[ \Delta T_b = K_b \cdot m \]
\[ 0.80 = 0.52 \cdot \left(\frac{\frac{12.5}{M}}{0.185}\right) \]

**Step 4: Rearrange the equation to solve for M.**
Rearranging gives:
\[ 0.80 = \frac{0.52 \cdot 12.5}{M \cdot 0.185} \]
Multiplying both sides by \(M \cdot 0.185\):
\[ 0.80 \cdot M \cdot 0.185 = 0.52 \cdot 12.5 \]
Now, solve for \(M\):
\[ M = \frac{0.52 \cdot 12.5}{0.80 \cdot 0.185} \]

**Step 5: Calculate the value of M.**
Calculating the right-hand side:
\[ M = \frac{6.5}{0.148} \approx 43.92 \, \text{g/mol} \]

### Final Answer:
The molar mass of the non-electrolyte substance is approximately **43.92 g/mol**.