A solution containing 1.8 g of a compound (empirical formula `CH_(2)O`) in 40 g of water is observed to freeze at `-0.465^(@)`C. The molecules formulea of the compound is (`K_(f)` of water =1.86kg K`mol^(-1)`):

Options

A. `C_(2)H_(4)O_(2)`
B. `C_(3)H_(6)`
C. `C_(4)H_(8)O_(4)`
D. `C_(6)H_(12)O_(6)`

Answer

To determine the molecular formula of the compound with an empirical formula of CH₂O, we will follow these steps:

### Step 1: Calculate the Freezing Point Depression
The freezing point depression (ΔTf) is given as -0.465°C. We will take the absolute value for calculations:
\[ \Delta T_f = 0.465 \, \text{°C} \]

### Step 2: Use the Freezing Point Depression Formula
The formula relating freezing point depression to molality is:
\[ \Delta T_f = K_f \times m \]
where:
- \( K_f \) is the cryoscopic constant of the solvent (water), given as 1.86 kg K/mol.
- \( m \) is the molality of the solution.

### Step 3: Rearrange the Formula to Find Molality
Rearranging the formula to solve for molality \( m \):
\[ m = \frac{\Delta T_f}{K_f} \]

### Step 4: Substitute the Values
Substituting the known values into the equation:
\[ m = \frac{0.465}{1.86} \approx 0.25 \, \text{mol/kg} \]

### Step 5: Calculate the Number of Moles of Solute
Molality is defined as moles of solute per kg of solvent. We have 40 g of water, which is 0.04 kg. Therefore:
\[ \text{moles of solute} = m \times \text{kg of solvent} = 0.25 \, \text{mol/kg} \times 0.04 \, \text{kg} = 0.01 \, \text{mol} \]

### Step 6: Calculate the Molar Mass of the Solute
We know the mass of the solute (the compound) is 1.8 g. The molar mass (M) can be calculated using the formula:
\[ M = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{1.8 \, \text{g}}{0.01 \, \text{mol}} = 180 \, \text{g/mol} \]

### Step 7: Determine the Molecular Formula
The empirical formula of the compound is CH₂O. The molar mass of the empirical formula is:
\[ \text{Molar mass of CH}_2\text{O} = 12 + (2 \times 1) + 16 = 30 \, \text{g/mol} \]
To find the molecular formula, we divide the molar mass of the compound by the molar mass of the empirical formula:
\[ \text{n} = \frac{\text{Molar mass of compound}}{\text{Molar mass of empirical formula}} = \frac{180 \, \text{g/mol}}{30 \, \text{g/mol}} = 6 \]
Thus, the molecular formula is:
\[ \text{C}_6\text{H}_{12}\text{O}_6 \]

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

---