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Solutions:- Part 2 – Preparation of solutions (Molar, Normal) and Dilution

Solutions:- Part 2 – Preparation of solutions (Molar, Normal) and Dilution
September 27, 2020Chemical pathologyLab Tests
  • These are the formulas for the preparation of various solutions.
  • Elaborated with examples.

Molar solution

  • It contains one mole of solute in a solution making equal to one liter.
  • Molar solution = Molecular weight in gram / liter in the solution.
  • Example: I molar solution of sodium chloride (NaCl).

Sodium atomic weight = 23

Chloride atomic weight  = 35.5

Total molecular weight = 58.5 gram / mol

Now dissolve 58.5 grams of NaCl in distilled water and make the solution to one liter.

Formation of Molar solution

Formation of Molar solution

Normal solution

  • The normal solution is defined as the gram equivalent weight per liter of the solution.
    • Normal solution  =  gram equivalent weight of solute/liter of the solution.
  • These solutions are expressed as N.
    • Gram equivalent weight = molecular weight/valency

Example

To make 1 N sodium chloride solution

  • The molecular weight of NaCl is 58.5
  • Gram equivalent weight of NaCl =  molecular weight / 1 (valency)
    • So dissolve 58.5 grams of NaCl in distilled water and makeup to one liter.

Percent solution

  1. This is per hundred part of the total solution.
  2. There are three possibilities of a percent solution.
    1. Weight/weight is this a percentage of solute in 100 grams of final solution equal to solute + solvent.
      1. e.g. 5 grams of NaCl dissolved in 95 grams of water which is around 95 mL.
    2. Weight/volume e.g. 5 grams of NaCl dissolved in water and the volume is made 100 ml is called a 5% solution of NaCl.
    3. Volume/volume composed of two solutions. e.g. if we take 5 mL of acid and dilute to 100 mL of water will be a 5% solution of that acid.

Dilution

  1. This procedure is very common to prepare the dilution of the serum where there is a high concentration of chemicals like urea in blood if it is above 300 mg/dL.
  2. If we make a dilution of serum like this:
    1. Serum = 1 ml
    2. Diluting fluid 4 mL
    3. This will be a dilution of 1:5 (1+4 =5).
Dilution 1:5 Solution Formation.

Dilution 1:5 Solution Formation.


Possible References Used
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