The diagram below illustrates an osmotic pressure experiment involving two solutions, 0.10 M aqueous NaCl and 0.15 M aqueous glucose (C6H12O6), at the same temperature. The semipermiable barrier only allows water to pass through. At the start (left diagram) equal volumes of the two solutions were placed in the apparatus. After a while the result on the right diagram was seen. Which of the two solutions was added on the left side of the apparatus? O impossible to tell O 0.15 M glucose O 0.10 M NaCl 00

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### Understanding Osmosis with NaCl and Glucose Solutions

The diagram below illustrates an osmotic pressure experiment involving two solutions: 
* 0.10 M aqueous NaCl 
* 0.15 M aqueous glucose (C₆H₁₂O₆)

Both solutions are at the same temperature. The setup involves a U-shaped tube with a semipermeable membrane in the middle, which only allows water to pass through.

#### Experimental Procedure
1. At the start (illustrated in the left diagram), equal volumes of the two solutions are introduced into the two arms of the U-tube.
2. Over time, the water movement due to osmosis results in a change observed in the right diagram.

#### Observations
- In the left diagram, the liquid levels on both sides of the U-tube are equal.
- In the right diagram, there is an unequal distribution of liquid levels, indicating osmotic movement of water from one side to the other.

#### Analysis
The key question is to determine which solution was added to the left side of the apparatus. Given the osmotic principles:
- NaCl, being an ionic compound, dissociates into two particles (Na⁺ and Cl⁻). Thus, 0.10 M NaCl produces an effective concentration of 0.20 M particles.
- Glucose, a non-electrolyte, remains as individual molecules, so 0.15 M glucose remains 0.15 M in particle concentration.

The side with the higher osmotic pressure (more particles) will draw water towards it. Since 0.20 M (from NaCl) > 0.15 M (from glucose), the side with NaCl will attract water more effectively, leading to a higher liquid level on that side.

Based on the diagrams:
- The higher level in the left side of the U-tube in the right diagram indicates that 0.10 M NaCl was added to the left side initially. 

#### Conclusion
The appropriate option to select is:

- ⭛ 0.10 M NaCl
Transcribed Image Text:### Understanding Osmosis with NaCl and Glucose Solutions The diagram below illustrates an osmotic pressure experiment involving two solutions: * 0.10 M aqueous NaCl * 0.15 M aqueous glucose (C₆H₁₂O₆) Both solutions are at the same temperature. The setup involves a U-shaped tube with a semipermeable membrane in the middle, which only allows water to pass through. #### Experimental Procedure 1. At the start (illustrated in the left diagram), equal volumes of the two solutions are introduced into the two arms of the U-tube. 2. Over time, the water movement due to osmosis results in a change observed in the right diagram. #### Observations - In the left diagram, the liquid levels on both sides of the U-tube are equal. - In the right diagram, there is an unequal distribution of liquid levels, indicating osmotic movement of water from one side to the other. #### Analysis The key question is to determine which solution was added to the left side of the apparatus. Given the osmotic principles: - NaCl, being an ionic compound, dissociates into two particles (Na⁺ and Cl⁻). Thus, 0.10 M NaCl produces an effective concentration of 0.20 M particles. - Glucose, a non-electrolyte, remains as individual molecules, so 0.15 M glucose remains 0.15 M in particle concentration. The side with the higher osmotic pressure (more particles) will draw water towards it. Since 0.20 M (from NaCl) > 0.15 M (from glucose), the side with NaCl will attract water more effectively, leading to a higher liquid level on that side. Based on the diagrams: - The higher level in the left side of the U-tube in the right diagram indicates that 0.10 M NaCl was added to the left side initially. #### Conclusion The appropriate option to select is: - ⭛ 0.10 M NaCl
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