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

# Solutions

### Definition of solution

- It is a mixture of liquid where the minor component is solute and is dissolved in the major component is solvent.
- This solute and solvent are uniformly distributed.

- These are the formulas for the preparation of various solutions

**Molar solution**

- It contains one mole as (molecular weight) of solute in a solution (solvent) making it 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.

## Normal solution

- The normal solution is defined as the gram equivalent weight per liter of the solution (solvent).
- Normal solution = gram equivalent weight of solute/liter of the solution (solvent) = Eq.wt/L.

- These solutions are expressed as N.
- Gram equivalent weight = Gram molecular weight/valency.

- Example of Gram equivalent weight e.g NaCl
- NaCl gram molecular weight = 58.5 g
- Valency =1
- 58.5/1 = 58.5 gram equivalent weight.

#### Example

**To make a 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.
**Dissolve 58.5 grams of NaCl in distilled water to make one liter.**

## Percent solution

- This is per hundred part of the total solution.
- There are three possibilities for a percent solution.
**Weight/weight:**- It is a percentage of solute in 100 grams of final solution equal to solute + solvent.
- e.g.For the 5% solution take 5 grams of NaCl dissolved in 95 grams of water which is around 95 mL.

**Weight/volume:**- 5 grams of NaCl dissolved in water and the volume is made 100 ml is called a 5% solution of NaCl.

**Volume/volume:**- It is composed of two solutions. e.g. if we take 5 mL of acid and dilute it to 100 mL of water will be a 5% solution of that acid.

## Dilution

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

- This dilution can be made from the stronger solution by this formula:

**Example dilution of sodium hydroxide 1:**- To make 250 mL of sodium hydroxide solution 0.25 mol/L from a solution of 0.4 mol/L solutions.
- C = o.25 mol/L
- V = 250 mL
- S = 0.4 mol/L
- Calculation = 0.25 x 250 / 0.4 = 156.25 mL

- To make 250 mL of sodium hydroxide solution 0.25 mol/L from a solution of 0.4 mol/L solutions.
**Example dilution of HCL 2:**- Make 500 ml of HCL acid , 0.01 mol/L from a 1 mol/L acid.
- C = 0.01
- V = 500
- S = 1
- Calculation = 0.01 x 500 / 1 = 5 mL of HCL acid.

- Make 500 ml of HCL acid , 0.01 mol/L from a 1 mol/L acid.
**Example for dilution of the body fluids:**- To make a 5 ml of 1 in 10 dilutions of the serum.
- C = 1:10
- Volume = 5 mL
- S = 1
- Calculations = 1/10 x 5 / 1 = 0.5 mL of the serum and 4.5 mL of the saline = 0.5 : 4.5 = 1:10 dilution

- To make a 5 ml of 1 in 10 dilutions of the serum.