2.i : Conversion of mg to ppm

By taking the RMM of the components of the chemical, it is easy to calculate how much of each will be in a 1M solution. A 1ppm solution of any single element will contain 0.001g of it per dm3, the same can be said for a 0.1 and 0.01ppm solutions. By finding these out for all of the components, adding the results together (1*1ppm + 2*0.1ppm + 7*0.01ppm for each), the mass of the solid required for the 1.27ppm solution is obtained.

Below is for all of the compounds used in my experiments. Note the K2Cr2O7 is not the same value as for the printed value. This is due to errors in values stated by the chemical companies. I have calculated them using the values given by IUPAC in 1969 (based on 12C).

Potassium Chromate 139.0925 1.27ppm Total
Symbol RMM 1M 1ppm 0.1ppm 0.01ppm (g) (g dm-3)
K 39.0983 39.0983 0.003248
Cr 51.996 51.996 0.002442
O3 47.9982 47.9982 0.002650 0.008340

2.ii : Sulphuric acid, 0.18 and 3 mol dm-3

For both of these, the concentrated acid is used first. This must be handled with great care. All volumes of this must be pipetted out in perfectly dry pipettes.

0.18 mol dm-3 : Use 10cm3 acid, dilute to a final volume of 1dm3.

3.0 mol dm-3 : Use 42cm3 acid, dilute to a final volume of 250cm3.

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2.iii : 1,5 Diphenyl carbazide (DiPC)

The method says to use 0.5g diluted to 200cm3 in propanone. By calculating how much DiPC is in each 50cm3, an easier value (for dilution) of 0.625g diluted to 250cm3 is obtained.

This solution must be made in a fume cupboard due to the solvent being propanone. It must also be stored away from light.

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2.iv : l-Ascorbic acid

0.2g was diluted in 100cm3 water.

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