Most community water supplies have 0.5 ppm of chlorine added for purification. What mass of chlorine must be added to 100.0 L of water to achieve this level?
Sun, 2008-07-20 22:02
i really need help with this question i dont understand how to do it because it is completely different from the example in my book!!!
Mon, 2008-07-21 13:09
The formula for parts per million is
ppm = mass of solute / mass of solution x 1,000,000
In this case you are told that you have a 100.0 L of water (the solution). Since the solution is essentially water, the density of the solution is very, very close to 1.00 g/ml.
And 100.0 L is equivalent to 100,000 mL. If we use the density, we can find the mass of the solution:
100,000 mL x 1.00 g/ml = 100,000 g
Now plug what we know into the equation for parts per million:
.5 ppm = x grams of solute / 100,000 g solution x 1,000,000
Solving for x yields x grams of solute = .5 x 100,000/ 1,000,000
x grams of solute = .5 x .1 = .05 grams of solute (chlorine)
Mon, 2008-07-21 14:58
basically if you can separate the magnitude of your units by 1000000 you will have ppm
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