Х = 0,355 ´ A х 5. Conclusion. SOLUTION. Х = 0,355 ´ A х 5

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Х = 0, 355 ´ A х 5

In 1 flask 0, 1 х 5 х 0, 355 = 0, 17 mg of chlorine on 1 litre

In 2 flask 0, 2 х 5 х 0, 355 = 0, 35 mg of chlorine on 1 litre

In 3 flasks 0, 4 х 5 х 0, 355 = 0, 71 mg of chlorine on 1 litre

We select the flask № 2, where residual chlorine - limits of norm.

N = 0, 3 - 0, 5 mg of chlorine on 1 litre.

In 2 flask we were adding 0, 2 mls of Sodium hyposulphitum ( 4 drops )

5). The amount of dry lime chloride in grammes necessary for a decontamination 1 cubic meter water is determined under the formula:

В х 5 х 1000 0, 2 х 5 х 1000

X = ------------------ = -------------------- = 10gr

100 100

Conclusion

The amount of dry lime chloride for a decontamination 1 cubic meter water = 10 grammes

Example 2

In infectious hospital it is necessary to disinfect water. After that water is come down into urban sewerage. It was titrated 200 milliliters of water in 3 flasks. In 1 flask we were adding 2 drops 0, 01 n Solution of Sodium hyposulphitum, in 2 flask - 4 drops and in 3 flask - 8 drops. What is necessary quantity of dry lime chloride for a decontamination 1000 м3 of water in infectious hospital?

SOLUTION

1). 1 ml 0, 01 n Solution of Sodium hyposulphitum contain 20 drops. In 1 flask we were add 2 drops

1 ml - 20 drops

Х ml - 2 drops

1 х 2

Х = -------- = 0, 1 mls of Sodium hyposulphitum

2). 1 ml 0, 01 n Solution of Sodium hyposulphitum contain 20 drops. In 2 flask we were add 4 drops

1 ml - 20 drops

Х ml - 4 drops

1 х 4

Х = -------- = 0, 2 mls of Sodium hyposulphitum

3). 1 ml 0, 01 n Solution of Sodium hyposulphitum contain 20 drops. In 3 flask we were add 8 drops

1 ml - 20 drops

Х ml - 8 drops

1 х 8

Х = -------- = 0, 4 mls of Sodium hyposulphitum

4). Residual quantity of chlorine in each flask determine under the formula:

Х = 0, 355 ´ A х 5

In 1 flask 0, 1 х 5 х 0, 355 = 0, 17 mg of chlorine on 1 litre

In 2 flask 0, 2 х 5 х 0, 355 = 0, 35 mg of chlorine on 1 litre

In 3 flasks 0, 4 х 5 х 0, 355 = 0, 71 mg of chlorine on 1 litre

We select the flask № 2, where residual chlorine - limits of norm.

N = 0, 3 - 0, 5 mg of chlorine on 1 litre.

In 2 flask we were adding 0, 2 mls of Sodium hyposulphitum ( 4 drops )

5). The amount of dry lime chloride in grammes necessary for a decontamination 1 cubic meter water is determined under the formula:

В х 5 х 1000 0, 2 х 5 х 1000

X = ------------------ = -------------------- = 10gr

100 100

6) The amount of dry lime chloride in grammes necessary for a decontamination 1000 м3 of water = 10gr x 1000 м3 = 10000 gr

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