The empirical formula of a non-electrolyte is `CH_(2)O`. A solution containing 3 g `L^(-1)` of the compound exerts the same osmotic pressure as that of 0.05 M glucose solution. The molecules formula of the compound is :

Options

A. (a) `CH_(2)O`
B. (b) `C_(2)H_(4)O_(2)`
C. (c) `C_(4)H_(8)O_(4)`
D. (d) `C_(3)H_(6)O_(3)`

Answer

To solve the problem, we need to find the molecular formula of a compound with the empirical formula \( CH_2O \) based on the information provided about its osmotic pressure compared to a glucose solution.

### Step-by-Step Solution:

1. **Understanding Osmotic Pressure**:
The osmotic pressure (\( \pi \)) of a solution can be calculated using the formula:
\[ \pi = iCRT \]
where:
- \( i \) = Van't Hoff factor (which is 1 for non-electrolytes),
- \( C \) = concentration of the solution in molarity (M),
- \( R \) = universal gas constant,
- \( T \) = temperature in Kelvin.

2. **Setting Up the Equation**:
Since both solutions exert the same osmotic pressure, we can set up the equation:
\[ \pi_{\text{non-electrolyte}} = \pi_{\text{glucose}} \]
This leads to:
\[ i_1 C_1 R T = i_2 C_2 R T \]
Given that \( i_1 = i_2 = 1 \) (both are non-electrolytes), we can simplify this to:
\[ C_1 = C_2 \]

3. **Finding the Concentration of the Non-Electrolyte**:
We know the concentration of glucose is \( 0.05 \, M \), thus:
\[ C_1 = 0.05 \, M \]

4. **Relating Concentration to Mass**:
The concentration in terms of mass can be expressed as:
\[ C = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)} \times \text{volume (L)}} \]
Given that the mass of the compound is \( 3 \, g/L \), we can express this as:
\[ 0.05 = \frac{3}{M} \]
where \( M \) is the molar mass of the compound.

5. **Calculating Molar Mass**:
Rearranging the equation gives:
\[ M = \frac{3}{0.05} = 60 \, g/mol \]

6. **Finding the Molecular Formula**:
The empirical formula \( CH_2O \) has a molar mass calculated as follows:
\[ \text{Molar mass of } CH_2O = 12 \, (C) + 2 \times 1 \, (H) + 16 \, (O) = 30 \, g/mol \]
To find the molecular formula, we divide the molar mass of the compound by the molar mass of the empirical formula:
\[ n = \frac{M}{\text{molar mass of } CH_2O} = \frac{60}{30} = 2 \]
Therefore, the molecular formula is:
\[ (CH_2O)_2 = C_2H_4O_2 \]

7. **Final Answer**:
The molecular formula of the compound is \( C_2H_4O_2 \).