Experiment Name: Formula related to textile wet processing.
Rule:-1: If material weight is "x" gram and M:L (Material : Liquor) is 1:N then total amount of liquor = x X n = ml or cc
Let
material weight = 8gm
M:L = 1:15
Total liquor =8 X 15
= 120mL
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,
material weight = 10gm
salt = 1%
stock solution = 1%
required amount of chemical
== (10X1%)/1% = 10mL
Rule-3:
If chemical amount recipe is expressed as gm/litre, then required amount of chemical,
==(Total liquor X chemical amount in gm/Litre ) / (Stock Solution(%)X1000)
Example :
Let
Fabric or material weight = 10gm
M : L = 1:10
Soda - Lime = 25gm/Litre
stock solution = 1%
Total liquor = 10 X 10 = 100mL
Soda Lime Required
= (100 X 25) / (1000 X 1%)
= 250 mL
Rule-4:
Calculation of additional water or initial water
amount of initial water = amount of total liquor - (Chemical 1 + chemical 2 + .......)
Example :
Let
Material weight = 5gm
M:L = 1:20
Total Liquor = 5 X 20 =100mL
Chemical - 1 = 10ml
Chemical -2 = 5ml
Initial Water = 100-(10+5)
= 100-15
=85mL
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.
Rule-6:
For the preparation of "y%" stock solution of a liquid chemical compound. we need to take (100-Y) mL distilled water and then add "y" mL of that liquid chemical to make 100 mL stock solution.
Example:
Let
We have to make 4% stock solution of HCl.
We need (100-4)=96 mL of distilled water and we have to add 4 ml HCl on 96 mL water.
