Question

When blood cells are placed in 1% (w/v) NaCl aqueous solution:

• A. Cell will burst
• B. Cell will shrink
• C. Cell will swell
• D. Cell remains as such

Hint:

Key Exam Tip:
When dealing with osmosis and cell behavior:
• Hypertonic Solution: Higher solute concentration outside the cell $\implies$ water moves out $\implies$ cell shrinks (crenation).
• Hypotonic Solution: Lower solute concentration outside the cell $\implies$ water moves in $\implies$ cell swells and may burst (hemolysis).
• Isotonic Solution: Equal solute concentration $\implies$ no net water movement $\implies$ cell maintains shape. For red blood cells, 0.9% NaCl is approximately isotonic.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem relates to the phenomenon of osmosis and the tonicity of solutions.
Osmosis is the net movement of solvent molecules through a semi-permeable membrane from a region of lower solute concentration to a region of higher solute concentration.
Human red blood cells (RBCs) are isotonic with a \( 0.9% \) (w/v) \( NaCl \) solution.
Step 2: Detailed Explanation:
When blood cells are placed in a \( 1% \) \( NaCl \) solution, the external concentration (\( 1% \)) is higher than the internal concentration (\( 0.9% \)).
This makes the \( 1% \) \( NaCl \) solution a hypertonic solution relative to the blood cell.
To equalize the concentration, water flows out of the cell through the cell membrane into the surroundings.
This process is specifically known as exosmosis.
Step 3: Final Answer:
The loss of water from the cell leads to a decrease in internal pressure and volume, causing the cell to shrink (crenation).