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 calculate the molar mass of the non-electrolyte substance based on the given data, we will use the formula for boiling point elevation:

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

where:
- \(\Delta T_b\) = boiling point elevation (in K)
- \(K_b\) = ebullioscopic constant (in K kg mol\(^{-1}\))
- \(m\) = molality of the solution (in mol/kg)

### Step 1: Identify the given values
- Mass of the solute (non-electrolyte substance), \(W_B = 12.5 \, \text{g}\)
- Mass of the solvent (water), \(W_A = 185 \, \text{g}\)
- Boiling point elevation, \(\Delta T_b = 0.80 \, \text{K}\)
- \(K_b = 0.52 \, \text{K kg mol}^{-1}\)

### Step 2: Convert mass of the solvent to kg
Since molality is defined as moles of solute per kg of solvent, we need to convert the mass of water from grams to kilograms:

\[ W_A = 185 \, \text{g} = 0.185 \, \text{kg} \]

### Step 3: Calculate molality (m)
Molality \(m\) is calculated using the formula:

\[ m = \frac{n_B}{W_A} \]

where \(n_B\) is the number of moles of solute. The number of moles can be expressed as:

\[ n_B = \frac{W_B}{M_B} \]

where \(M_B\) is the molar mass of the solute. Therefore, we can express molality as:

\[ m = \frac{W_B}{M_B \cdot W_A} \]

### Step 4: Substitute molality into the boiling point elevation equation
Substituting the expression for molality into the boiling point elevation equation gives:

\[ \Delta T_b = K_b \cdot \frac{W_B}{M_B \cdot W_A} \]

### Step 5: Rearrange to solve for molar mass \(M_B\)
Rearranging the equation to solve for \(M_B\):

\[ M_B = \frac{K_b \cdot W_B}{\Delta T_b \cdot W_A} \]

### Step 6: Substitute the known values into the equation
Substituting the known values into the equation:

\[ M_B = \frac{0.52 \, \text{K kg mol}^{-1} \cdot 12.5 \, \text{g}}{0.80 \, \text{K} \cdot 0.185 \, \text{kg}} \]

### Step 7: Calculate the numerator and denominator
Calculating the numerator:

\[ 0.52 \cdot 12.5 = 6.5 \, \text{g kg mol}^{-1} \]

Calculating the denominator:

\[ 0.80 \cdot 0.185 = 0.148 \, \text{kg} \]

### Step 8: Final calculation for molar mass
Now, substituting back into the equation:

\[ M_B = \frac{6.5}{0.148} \approx 43.92 \, \text{g/mol} \]

### Conclusion
The molar mass of the non-electrolyte substance is approximately \(43.91 \, \text{g/mol}\).

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