# Fertilisers (Sampling and Analysis) Regulations (Northern Ireland) 1996, Paragraph 8: EXPRESSION OF RESULTS


1 ml of 0.02 M potassium permanganate solution corresponds to 1.099 mg of manganese (Mn). The percentage of manganese in the fertiliser is given by:
Mn(%)where=(xb-xs)×0.1099×———Va×M where:
xb is the volume in ml of the permanganate used for the blank;
xs is the volume in ml of the permanganate used for the test sample;
V is the volume in ml of the extract solution in accordance with Methods 26a and 26b;
a is the volume in ml of the aliquot portion taken from the extract (6.2) or (6.3);
M is the mass in g of the test sample.

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Source: https://www.legislation.gov.uk/nisr/1996/513/schedule/2/part/I/chapter/26i./paragraph/8 (the official version)
Page: https://openlegislation.uk/fertilisers-sampling-and-analysis-regulations-northern-ireland-1996/schedule-2/part-I/chapter-26i./paragraph-8
Contains public sector information licensed under the Open Government Licence v3.0. Source: legislation.gov.uk, The National Archives.