A forensic chemist tests a water sample for cyanide. To do so, the sample is diluted by pipetting 1 mL into an empty 100 mL volumetric flask, which is then filled to the mark with deionized water. If the measured concentration was found to be 0.396 mg/L, what was the concentration of the original sample. (answer in ppm.)
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calculate the original concentration using the equation:
C1V1 = C2V2
C1*1 = 0.396*100
C1 = 39.6/1
C1 = 39.6 mg/L
mg/L = ppm
Therefore the original solution concentration was 39.6ppm.
C1V1 = C2V2
C1*1 = 0.396*100
C1 = 39.6/1
C1 = 39.6 mg/L
mg/L = ppm
Therefore the original solution concentration was 39.6ppm.
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If the concentration of CN- ions in 1 litre was 0.396 mg then the number of milligrams of CN- ions in 101ml will be (0.396/1000) x 101 = 0.039996 mg
Now, number of mg of CN- in 10^6 g water = 0.039996/101 x 10^6 = 396 ppm
Now, number of mg of CN- in 10^6 g water = 0.039996/101 x 10^6 = 396 ppm
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I think mg/L is ppm so in the volumetric flask is 0.396 ppm...and (100/1) x 0.396 ppm is 39.6 ppm, the sample concentration is 39.6 ppm