Two beakers, one contaning `20ml` of a `0.05M` aqueous solution of a non volatile, non electrolyte and the other, the same volume of `0.03M` aqueous solution of `NaCl`, are placed side by side in a closed enclose. What are the volumes in the two beakers when equilibrium is attained ? Volume of the solution in the first and second beaker are respectively.

Options

A. `21.8mL` and `18.2mL`
B. `18.2mL` and `21.8mL`
C. `20mL` and `20mL`
D. `17.1mL` and `22.9mL`

Explanation

To solve the problem, we need to determine the volumes in two beakers when equilibrium is attained after placing them side by side in a closed enclosure. One beaker contains a non-volatile, non-electrolyte solution, and the other contains a NaCl solution.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Beaker 1: 20 ml of 0.05 M non-volatile, non-electrolyte solution.
- Beaker 2: 20 ml of 0.03 M NaCl solution.

2. **Calculate Moles of Solute in Beaker 1:**
- Moles of solute (non-volatile, non-electrolyte) = Molarity × Volume (in liters)
- Volume in liters = 20 ml = 20/1000 = 0.020 L
- Moles of solute = 0.05 M × 0.020 L = 0.001 moles.

3. **Calculate Moles of Solute in Beaker 2:**
- NaCl dissociates into two ions (Na⁺ and Cl⁻), so the effective concentration of solute particles is doubled.
- Moles of solute (NaCl) = Molarity × Volume (in liters) × 2 (for dissociation)
- Moles of solute = 0.03 M × 0.020 L × 2 = 0.0012 moles.

4. **Determine the Direction of Water Movement:**
- Since Beaker 2 has a higher concentration of solute particles (0.0012 moles) compared to Beaker 1 (0.001 moles), water will move from Beaker 1 to Beaker 2 until equilibrium is reached.

5. **Set Up the Equation for Equilibrium:**
- Let \( V \) be the volume of water that flows from Beaker 1 to Beaker 2.
- At equilibrium:
- Volume in Beaker 1 = \( 20 - V \) ml
- Volume in Beaker 2 = \( 20 + V \) ml
- Concentration in Beaker 1:
\[ \text{Concentration} = \frac{0.001}{20 - V} \]
- Concentration in Beaker 2:
\[ \text{Concentration} = \frac{0.0012}{20 + V} \]

6. **Set the Concentrations Equal for Equilibrium:**
\[ \frac{0.001}{20 - V} = \frac{0.0012}{20 + V} \]

7. **Cross-Multiply to Solve for \( V \):**
\[ 0.001(20 + V) = 0.0012(20 - V) \]
\[ 0.02 + 0.001V = 0.024 - 0.0012V \]
\[ 0.001V + 0.0012V = 0.024 - 0.02 \]
\[ 0.0022V = 0.004 \]
\[ V = \frac{0.004}{0.0022} \approx 1.818 \text{ ml} \approx 1.8 \text{ ml} \]

8. **Calculate Final Volumes in Each Beaker:**
- Volume in Beaker 1 = \( 20 - V = 20 - 1.8 = 18.2 \) ml
- Volume in Beaker 2 = \( 20 + V = 20 + 1.8 = 21.8 \) ml

### Final Answer:
The volumes in the two beakers when equilibrium is attained are:
- Beaker 1: **18.2 ml**
- Beaker 2: **21.8 ml**