A certain non-volatile electrolyte contain 40% carbon, 6.7% hydrogen and 53.3% oxygen.An aqueous solution containing 5% by mass of the solute boils at `100.15^(@)`C. Determine molecular formula of the compound(`K_(b) =0.51^(@)C//m`):

Options

A. HCHO
B. `CH_(3)OH`
C. `C_(2)H_(5)OH`
D. `C_(6)H_(12)O_(6)`

Answer

To determine the molecular formula of the compound from the given data, we will follow these steps:

### Step 1: Calculate the elevation in boiling point (ΔTb)
The boiling point elevation can be calculated using the formula:
\[ \Delta T_b = K_b \times m \]
Where:
- \( \Delta T_b \) = boiling point elevation
- \( K_b \) = ebullioscopic constant (given as 0.51 °C/m)
- \( m \) = molality of the solution

### Step 2: Find the boiling point elevation
The boiling point of pure water is 100 °C, and the boiling point of the solution is 100.15 °C. Therefore:
\[ \Delta T_b = 100.15 °C - 100 °C = 0.15 °C \]

### Step 3: Rearranging the formula to find molality
We can rearrange the formula to find molality:
\[ m = \frac{\Delta T_b}{K_b} \]
Substituting the known values:
\[ m = \frac{0.15 °C}{0.51 °C/m} \approx 0.2941 \, m \]

### Step 4: Calculate the mass of the solvent
Given that the solution is 5% by mass of the solute, in 100 g of solution, the mass of the solute is 5 g. Therefore, the mass of the solvent (water) is:
\[ \text{Mass of solvent} = 100 \, g - 5 \, g = 95 \, g \]

### Step 5: Calculate the number of moles of solute
Molality is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
Converting the mass of the solvent to kg:
\[ 95 \, g = 0.095 \, kg \]
Now, substituting the values into the molality equation:
\[ 0.2941 = \frac{\text{moles of solute}}{0.095} \]
Rearranging to find the moles of solute:
\[ \text{moles of solute} = 0.2941 \times 0.095 \approx 0.0279 \, moles \]

### Step 6: Calculate the molecular weight of the solute
We know that the mass of the solute is 5 g. Therefore, we can find the molecular weight (M) using the formula:
\[ M = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{5 \, g}{0.0279 \, moles} \approx 179.8 \, g/mol \]

### Step 7: Determine the composition of the compound
The compound consists of 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. We can calculate the mass of each element in 180 g of the compound (using the molecular weight we just calculated):
- Mass of carbon = 40% of 180 g = \( 0.40 \times 180 = 72 \, g \)
- Mass of hydrogen = 6.7% of 180 g = \( 0.067 \times 180 \approx 12.06 \, g \)
- Mass of oxygen = 53.3% of 180 g = \( 0.533 \times 180 \approx 96 \, g \)

### Step 8: Calculate the number of moles of each element
- Moles of Carbon (C):
\[ \text{Moles of C} = \frac{72 \, g}{12 \, g/mol} = 6 \, \text{moles} \]
- Moles of Hydrogen (H):
\[ \text{Moles of H} = \frac{12.06 \, g}{1 \, g/mol} \approx 12 \, \text{moles} \]
- Moles of Oxygen (O):
\[ \text{Moles of O} = \frac{96 \, g}{16 \, g/mol} = 6 \, \text{moles} \]

### Step 9: Write the empirical formula
From the calculations, we have:
- C: 6
- H: 12
- O: 6

Thus, the molecular formula of the compound is:
\[ \text{C}_6\text{H}_{12}\text{O}_6 \]

### Conclusion
The molecular formula of the compound is \( \text{C}_6\text{H}_{12}\text{O}_6 \).

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