1) Aldrin and dieldrin are pesticides that used to be allowed for the control of soil insects. In ontario, the maximum allowable total concentration of alrin plus dieldrin in drinking water is 0.7 ppb. If a 250mL sample of drinking water is found to contain 0.0001mg of aldrin and dieldrin, does the concentration exceed the standard? Explain.
I tried doing this and kept on getting the wrong answer please help me out
0.7 ppb = 0.7 microgram/L
1E-5mg Al+Diel/250 = 4E-5 mg/L. Convert to ug.
4E-5mg x (1000 ug/1 mg) = 4E-2 ug = 0.04ug.
So you have 0.04 ug/L.= 0.04 ppb
what is ug and microgram and al+diel...im in grade 11 have not learnt this yet..
To determine whether the concentration exceeds the standard, we need to convert the units and compare it to the maximum allowable concentration of 0.7 ppb (parts per billion).
First, let's convert the sample amount from milligrams (mg) to ppb. To do this, we need to consider that 1 ppb is equivalent to 1 microgram (µg) per liter (L).
Since we have a 250 mL sample, we need to convert it to liters:
250 mL = 0.25 L
Now, let's convert the amount of aldrin and dieldrin (0.0001 mg) to µg:
0.0001 mg = 100 µg (since 1 mg = 1000 µg)
Next, we divide the amount of aldrin and dieldrin (100 µg) by the volume of the sample (0.25 L):
100 µg ÷ 0.25 L = 400 µg/L
Finally, we have the concentration in µg/L, which can be directly compared to the standard of 0.7 ppb.
To convert µg/L to ppb, we move the decimal point three places to the left:
400 µg/L = 0.4 ppb
Since the calculated concentration of aldrin and dieldrin (0.4 ppb) is below the maximum allowable concentration of 0.7 ppb, the concentration does not exceed the standard.
Therefore, the sample of drinking water does not exceed the maximum allowable total concentration of aldrin plus dieldrin in Ontario.