please show all steps and calculations. 1. You are using RBC suspended in isoton

Question

please show all steps and calculations.

1. You are using RBC suspended in isotonic solutions of NaCl and Glucose, to avoid RBC damage. You need to prepare 500 mL of the NaCl solution and 250 mL of the Glucose solution.

a. How much NaCl and glucose do you need to weigh out to prepare these solutions?

b. What is the concentration of NaCl and glucose, in their respective solution, in %w/v, g/L and mM?

2. You prepare a 0.45 % NaCl solution.

a. Is this solution correct?

b. What is the osmolarity (mOsM) of this solution?

c. Tonicity-wise, how would you categorize this solution?

d. If you suspend RBC in this solution what would be the most likely outcome?

Explanation / Answer

Hi,
The isotonic solution of Nacl and glucose contains 0.9% salt and 5% glucose.
To make 500 Nacl solution, we need to dissolve (0.9 grams in 100 ml,)so in 500 ml ; 0.9 x 5 = 4.5 gms.
To make 250 ml of glucose , we need to dissolve (5 gm in 100 ml) = 5 x 2.5 = 12.5 gm.

concentration of Nacl %w/v = 4.5 *100/ 500 = 0.9%
in terms of g/L = 9 gm / L
in mM : 58.44 gm - 1l = 1M
9/58.44 = 0.154 M = 154mM.

2. 0.45% Nacl solution in not correct. It should be 0.9%.

The osmolarity of this solution = 154 mOsm/kg
Formula: NaCl dissociates completely in water to form Na+ ions and Cl ions.
Thus, each mole of NaCl becomes two osmoles in solution: one mole of Na
+ and one mole of Cl. A solution of 1 mol/L NaCl has an osmolarity of 2 Osmol/L.
No. of moles of Na in this solution = 154/2 = 77
Na osmolarity = ( 77/1 ) X (2/1) = 154
This is a hypotonic solution.
If we suspend RBC, it would swell and burst open.

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