Two beaker A and B present in a closed vessel. Beaker A contains 152.4 g aqueous soulution of urea, containing 12 g of urea. Beaker B contain 196.2 g glucose solution, containing 18 g of glocose. Both solution allowed to attain the equilibrium. Determine mass % of glocose in its solution at equilibrium allowed to attain the equilibrium :

Options

A. 6.71
B. 14.49
C. 16.94
D. 20

Answer

To solve the problem, we need to determine the mass percentage of glucose in its solution at equilibrium after the two solutions have been allowed to reach equilibrium. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the given data
- **Beaker A (Urea solution)**:
- Total mass of solution = 152.4 g
- Mass of urea = 12 g
- Mass of water = Total mass - Mass of urea = 152.4 g - 12 g = 140.4 g

- **Beaker B (Glucose solution)**:
- Total mass of solution = 196.2 g
- Mass of glucose = 18 g
- Mass of water = Total mass - Mass of glucose = 196.2 g - 18 g = 178.2 g

### Step 2: Calculate the number of moles of urea and water in Beaker A
- **Molecular weight of urea (NH₂CONH₂)** = 60 g/mol
- Number of moles of urea = Mass of urea / Molecular weight of urea = 12 g / 60 g/mol = 0.2 mol
- **Molecular weight of water (H₂O)** = 18 g/mol
- Number of moles of water = Mass of water / Molecular weight of water = 140.4 g / 18 g/mol = 7.8 mol

### Step 3: Calculate the mole fraction of urea in Beaker A
- Mole fraction of urea (X_Urea) = Moles of urea / (Moles of urea + Moles of water)
- X_Urea = 0.2 / (0.2 + 7.8) = 0.2 / 8 = 0.025

### Step 4: Calculate the number of moles of glucose and water in Beaker B
- **Molecular weight of glucose (C₆H₁₂O₆)** = 180 g/mol
- Number of moles of glucose = Mass of glucose / Molecular weight of glucose = 18 g / 180 g/mol = 0.1 mol
- Number of moles of water = Mass of water / Molecular weight of water = 178.2 g / 18 g/mol = 9.9 mol

### Step 5: Calculate the mole fraction of glucose in Beaker B
- Mole fraction of glucose (X_Glucose) = Moles of glucose / (Moles of glucose + Moles of water)
- X_Glucose = 0.1 / (0.1 + 9.9) = 0.1 / 10 = 0.01

### Step 6: Analyze the equilibrium condition
Since the mole fraction of urea is greater than that of glucose, the vapor pressure of the urea solution is higher. Therefore, water will move from the glucose solution to the urea solution until equilibrium is reached.

### Step 7: Let x moles of water be transferred from glucose to urea
- At equilibrium:
- Moles of water in Beaker A = 7.8 + x
- Moles of water in Beaker B = 9.9 - x

### Step 8: Set up the equation for mole fractions at equilibrium
- For urea:
\[ \text{X}_{\text{Urea}} = \frac{0.2}{0.2 + (7.8 + x)} \]
- For glucose:
\[ \text{X}_{\text{Glucose}} = \frac{0.1}{0.1 + (9.9 - x)} \]

At equilibrium, the mole fractions will equalize.

### Step 9: Solve for x
Setting the two mole fractions equal:
\[ \frac{0.2}{0.2 + (7.8 + x)} = \frac{0.1}{0.1 + (9.9 - x)} \]
Cross-multiplying and simplifying will yield the value of x. After solving, we find that \( x = 4 \) moles of water are transferred.

### Step 10: Calculate the new mass of glucose solution
- New mass of glucose solution = Original mass - Mass of water transferred
- Mass of water transferred = \( 4 \text{ moles} \times 18 \text{ g/mole} = 72 \text{ g} \)
- New mass of glucose solution = 196.2 g - 72 g = 124.2 g

### Step 11: Calculate the mass percentage of glucose at equilibrium
- Mass percentage of glucose = (Mass of glucose / Mass of glucose solution) × 100
- Mass percentage of glucose = (18 g / 124.2 g) × 100 = 14.49%

### Final Answer
The mass percentage of glucose in its solution at equilibrium is **14.49%**.

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