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 = (x b - x s) \times 0.1099 \times ——— \frac{V}{a \times M}

where:

x_b is the volume in ml of the permanganate used for the blank;
x_s 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.