The vapour pressure of a dilute aqueous solution of glucose is `750`mmHg at `373`K. The mole fraction of the solute is

Options

A. `(1)/(10)`
B. `(1)/(7.6)`
C. `(1)/(35)`
D. `(1)/(76)`

Answer

To find the mole fraction of glucose in a dilute aqueous solution given its vapor pressure, we can follow these steps:

### Step 1: Understand the given data
- The vapor pressure of the solution (P_s) = 750 mmHg
- The vapor pressure of pure water (P°_H2O) at 373 K = 760 mmHg

### Step 2: Apply Raoult's Law
According to Raoult's Law, the vapor pressure of a solution is given by:
\[ P_s = P°_{H2O} \cdot X_{H2O} \]
Where:
- \( P_s \) = vapor pressure of the solution
- \( P°_{H2O} \) = vapor pressure of pure solvent (water)
- \( X_{H2O} \) = mole fraction of the solvent (water)

### Step 3: Express the mole fraction of the solvent
The mole fraction of the solvent can be expressed in terms of the mole fraction of the solute (glucose):
\[ X_{H2O} = 1 - X_{glucose} \]
Where:
- \( X_{glucose} \) = mole fraction of the solute (glucose)

### Step 4: Substitute the mole fraction into Raoult's Law
Substituting \( X_{H2O} \) into the equation gives:
\[ P_s = P°_{H2O} \cdot (1 - X_{glucose}) \]

### Step 5: Rearrange the equation
Rearranging the equation to solve for \( X_{glucose} \):
\[ 750 = 760 \cdot (1 - X_{glucose}) \]

### Step 6: Solve for \( X_{glucose} \)
1. Divide both sides by 760:
\[ \frac{750}{760} = 1 - X_{glucose} \]
2. Calculate \( \frac{750}{760} \):
\[ \frac{750}{760} = 0.9868 \]
3. Rearranging gives:
\[ X_{glucose} = 1 - 0.9868 = 0.0132 \]

### Step 7: Convert to mole fraction form
To express this as a fraction:
\[ X_{glucose} = 0.0132 = \frac{1}{76} \]

### Conclusion
Thus, the mole fraction of glucose in the solution is:
\[ X_{glucose} = \frac{1}{76} \]