According to photo metric measurements, 1 drop of activated Sodium Chlorite ( 0.05 mL) produces about 0.4 mg ClO₂ in a solution; that is the equivalent of about 0.2 mg of H202.

If a 3% solution is used, this 0.2 mg of H202 would be contained in 0.007 mL, i.e. about 1/7th of a drop (3% x 0.2 mg / 100 mL = approx .0006 mL). That's fine.

But than who does this calculation is done(though similar to above), am failing to understand......

The calculation is as follows:

Thanks.

If a 3% solution is used, this 0.2 mg of H202 would be contained in 0.007 mL, i.e. about 1/7th of a drop (3% x 0.2 mg / 100 mL = approx .0006 mL). That's fine.

But than who does this calculation is done(though similar to above), am failing to understand......

The calculation is as follows:

**100 mL 3% H202 solution = 3000 milligrams H202**

x ml = 0.2 milligrams H202 ??

x – 0.007 ml ??x ml = 0.2 milligrams H202 ??

x – 0.007 ml ??

Thanks.

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