Formula related to textile wet processing

Experiment Name: Formula related to textile wet processing.

Rule 1: If material weight is "x" gram and M:L (Material : Liquor) is total amount of liquor = (x × n) = ml or cc (n = weight of liquor)

Example:
Let
M:L = 1:15,
Material weight = 8gm,
Total liquor = (8 × 15)
                    = 120 ml or cc

Rule 2: If chemical amount in recipe is expressed as percentage (%). Then required amount chemical = (Material Weight X Chemical amount in recipe) / stock solution (%).
Example:
Let
M:L = 1:15,
Material weight = 8gm,
Salt = 1%,
Stock solution = 1%,
Required amount of chemical $=\frac{10\times 1\%}{1\%}$
                                                  = 10 ml or cc.

Rule 3: If chemical amount recipe is expressed as gm/litre, then required amount of chemical,
$=\frac{Total liquor \times chemical amount in gm/Litre}{Stock Solution\%\times 1000}$
Example:
Let
M:L = 1:15,
Material weight = 8gm,
Soda Lime = 25gm/Litre.
Stock solution = 1%
Total liquor = 8 X 15 = 120 ml or cc.
Required Soda Lime $=\frac{120\times 25}{1000\times 1\%}$
                                    =300 ml or cc.

Rule 4: Calculation of additional water or initial water amount of initial water
= amount of total liquor - (Chemical 1 + chemical 2 + .......)
Example:
Let
M:L = 1:15,
Material weight = 8gm,
Total liquor = (8 × 15)
                    = 120 ml or cc.
Chemical 1 = 10 ml or cc.
Chemical 2 = 5 ml or cc.
Initial Water = {120 - (10 + 5)}
                     = 105 ml or cc.

Rule 5: For the preparation of "x%" stock solution of a solid chemical compound we need to take 100 mL of distilled water and then add "x" gm of that chemical to make 100 mL solution.
Example: If we want to make 5% stock solution of NaCl.
We have to dissolve 5gm of NaCl in 100 mL of distilled water.

Conclusion:

Our teacher and lab assistants are very much helpful to us. Their well teaching and instruction help us greatly to understand of equation of dyeing. I think this practical will be very helpful in my future career.