20 g of non-electrolyte, non-volatile solute `(C_(x)H_(2x)O_(6))` when dissolved in 100 gm water at `100^(@)C`, lowers the vapour pressure of solution by `(1)/(100)`th of the vapour pressure of pure water at this temperature. What is formula of the compound ?

Options

A. `C_(6)H_(12)O_(6)`
B. `C_(12)H_(24)O_(12)`
C. `C_(44)H_(88)O_(44)`
D. `C_(3)H_(6)O_(3)`

Answer

To solve the problem, we will use Raoult's Law, which relates the vapor pressure of a solution to the vapor pressure of the pure solvent and the mole fraction of the solvent in the solution.

### Step-by-Step Solution:

1. **Understanding the Given Data:**
- Mass of solute, \( W_2 = 20 \, \text{g} \)
- Mass of solvent (water), \( W_1 = 100 \, \text{g} \)
- Lowering of vapor pressure, \( \Delta P = \frac{1}{100} P_0 \) (where \( P_0 \) is the vapor pressure of pure water at \( 100^\circ C \))

2. **Using Raoult's Law:**
According to Raoult's Law, the relative lowering of vapor pressure is given by:
\[ \frac{\Delta P}{P_0} = \frac{n_2}{n_1 + n_2} \]
where \( n_1 \) is the number of moles of the solvent and \( n_2 \) is the number of moles of solute.

3. **Calculating Moles of Solvent:**
The molar mass of water (H₂O) is approximately \( 18 \, \text{g/mol} \).
\[ n_1 = \frac{W_1}{\text{Molar mass of water}} = \frac{100 \, \text{g}}{18 \, \text{g/mol}} \approx 5.56 \, \text{mol} \]

4. **Setting Up the Equation:**
From the problem, we know that:
\[ \frac{\Delta P}{P_0} = \frac{1}{100} \]
Thus, we can write:
\[ \frac{1}{100} = \frac{n_2}{n_1 + n_2} \]

5. **Substituting Values:**
Let \( n_2 = \frac{W_2}{M_2} \), where \( M_2 \) is the molar mass of the solute.
\[ \frac{1}{100} = \frac{\frac{20}{M_2}}{5.56 + \frac{20}{M_2}} \]

6. **Cross-Multiplying:**
Cross-multiplying gives:
\[ 5.56 + \frac{20}{M_2} = 100 \cdot \frac{20}{M_2} \]
Simplifying this:
\[ 5.56 M_2 + 20 = 2000 \]
\[ 5.56 M_2 = 1980 \]
\[ M_2 = \frac{1980}{5.56} \approx 356.25 \, \text{g/mol} \]

7. **Identifying the Compound:**
The formula of the compound is given as \( C_xH_{2x}O_6 \). We need to find \( x \) such that the molar mass equals approximately \( 360 \, \text{g/mol} \).
The molar mass of \( C_xH_{2x}O_6 \) can be calculated as:
\[ \text{Molar mass} = 12x + 2(2x) + 6(16) = 12x + 4x + 96 = 16x + 96 \]
Setting this equal to \( 360 \):
\[ 16x + 96 = 360 \]
\[ 16x = 264 \]
\[ x = \frac{264}{16} = 16.5 \]
Since \( x \) must be a whole number, we can round \( x \) to 12, leading us to \( C_{12}H_{24}O_6 \) which has a molar mass of \( 360 \, \text{g/mol} \).

### Final Answer:
The formula of the compound is \( C_{12}H_{24}O_{12} \).