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Postby mcvintosh » Wed Apr 26, 2006 4:46 am

could you help me calculate how much volume of 2500mg/l solution of chlorine at a rate of 3mg/l would be added to 3 million litres of water.

Postby Guest » Wed Apr 26, 2006 8:18 pm

Approach 1. For 3E6 L of solution:
3 mg/L x 3E6 L = 9000 g Cl

All the Cl comes from the bleach solution, at its concentration
9000 g / 2500 mg/L = 3600 L

Note this makes 3E6 L of total solution counting the 3600 L of bleach. only (3E6 - 3600) L of water were used.
3600 L Cl solution/ 2 996 400 L H2O x 3E6 L H2O = 3604 L

Approach 2
2500 mg/L / 3 mg/L = 833.333 x too strong
You want to dilute each part of Cl solution with 832.333 parts water (to make 833.333 parts total)
1E6/832.333 = 3604. L

Postby Guest » Wed Apr 26, 2006 8:20 pm

Anonymous wrote:1E6/832.333 = 3604. L

Oops. Last line should read "3E6/832.333 = 3604. L"
Math was right, typing was not.

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