Q. 9.70 - An Introduction to General, Organic, and Biological Chemistry

Question

Will a red blood cell undergo crenation, hemolysis, or no change in each of the following solutions?

a. $1\% (m/v)$ glucose

b. $2\% (m/v) NaCl$

c. $5\% (m/v)$ glucose

d. $0.1\% (m/v) NaCl$

Step-by-Step Solution

Answer

(part a) The following solution $1\% (m/v)$ glucose undergoes Hemolysis.

(part b) The following solution $2\% (m/v) NaCl$, The red blood cell will crenation.

(part c) The following solution $5\% (m/v)$ glucose. There is no change in this solution.

(part d) The following solution $0.1\% (m/v) NaCl$. This solution leads to Hemolysis.

Step1: Introduction (part a).

• A red blood cell maintains its quantity in an electrolyte as there is an equal stream of fluid into and out of cell. Isotonic solutions have included a $0.9\% (m/v) NaCl$solution and a $5\% (m/v)$ glucose solution.
• When a red blood cell is submersed in a hypotonic, water enters the cell and causes it to expand and burst. This is known as hemolysis. The solute concentration in a hypotonic solution is lower.
• When a red blood cell is positioned in a hypertonic solution, moisture escapes, causing the cell to shrink. This is known as crenation. A hypertonic has a higher concentration of solutes.

Step2: Given Information (part a).

Given:

$1\% (m/v)$ glucose.

Step3: Explanation (part a).

A $1\% (m/v)$ glucose solution does have a lower osmotic pressure than a $5\% (m/v)$ glucose solution. As just a result, a $1\% (m/v)$ glucose solution is hypotonic. As a result, a red blood cell will be hemolyzed.

Step4: Given Information (part b).

Given:

$2\% (m/v) NaCl$

Step5: Explanation (part b).

A $2\% (m/v) NaCl$ solution contains more solutes than a $0.9\% (m/v) NaCl$solution. As a result, a $2\% (m/v) NaCl$ solution is hypertonic. As a result, a red blood cell would then crenate.

Step6: Given Information (part c).

Given:

$5\% (m/v)$ glucose.

Step7: Explanation (part c).

A $5\% (m/v)$ sucrose solution is isotonic. Therefore, a red blood cell will remain unchanged.

Step8: Given Information (part d).

Given:

$0.1\% (m/v) NaCl$

Step9: Explanation (part d).

A $0.1\% (m/v) NaCl$ solution contains less solute than a $0.9\% (m/v) NaCl$ solution. As a result, a $0.1\% (m/v) NaCl$ solution is hypotonic. As a result, a red blood cell will be hemolyzed.