For any textile process we often need to calculate the amount of chemicals to be added in the process liquor and many a times we need to convert or dilute their concentrations in order to suit the required concentration.

__NOTE:__ Both ‘P’ & ‘C’ should be in same concentration units either ‘%’ or ‘gpl’. If different, for example Required conc. is in gpl and available conc. is in %, then we have to convert the unit first as shown below.

**For converting ‘%’ into ‘gpl’ the % is multiplied by 10**. For example to convert 5% into gpl, multiply it with 10,

i.e., 5% X 10 = 50 gpl

For 10%, 10% X 10 = 100 gpl

For 40%, 40% X 10 = 400 gpl

Now, **for converting ‘gpl’ into ‘%’ divide it by 10**. For example, to convert 10gpl into %, divide it by 10,

i.e., 10gpl / 10 = 1%

For 5gpl 5gpl / 10 = 0.5%

For 25gpl 25gpl / 10 = 2.5%

**DERIVATION #1 for Calculations based on Weight of Material**

Let the Required conc. = P %

Weight of material = W

Available conc. = C %

Total Volume of Liquor = V

Now,

for 100gms of material –> ‘P’ gms of chemical is required,

Therefore for ‘W’ gms of material –> W X P / 100 gms of chemical is required,

* ………….(by cross multiplication)*

Similarly,

‘C’ gms of chemical is present in –> 100 ml of solution

Therefore, (W X P / 100) gms of chemical is present in –> {(W X P / ~~100~~) * ~~100~~} / C ml of solution

*…………….(by cross multiplication)*

= (W X P / C) ml of solution

Hence Proved,** for ‘W’ gms of Material, (W X P / C) ml of solution is taken**

**DERIVATION #2 for Calculations based on Volume of Liquor**

Let the Required Conc. = P gpl

Weight of Material = W

Available conc. = C gpl

Total Volume of Liquor = V

Now,

‘P’ gpl means, P gms of chemical is present in –> 1000 ml of solution

i.e., for 1000 ml of solution –> ‘P’ gms of chemical is required

Therefore, for ‘V’ ml of solution –> (V X P / 1000) gm of chemical is required.

* ………….(by cross multiplication)*

Similarly,

‘C’ gpl means, C gm of chemical is present in –> 1000 ml of solution

Therefore, (V X P / 1000) gms of chemical is present in –> {(V X P / ~~1000~~) * ~~1000~~ / C}

* ………….(by cross multiplication)*

= (V X P / C) ml of solution

Hence Proved,** for ‘V’ ml of Liquor, (V X P / C) ml of solution is taken**