1. A gram of soil sample as received contains 15.00% water. After oven drying to remove the mixture, it was found to contain 16.00% K. Find the %K in the original sample as received.
16.00−15.00=1.0016.00-15.00=1.0016.00−15.00=1.00 %
%K=1.0016.00×100=6.25=\frac{1.00}{16.00}×100=6.25=16.001.00×100=6.25 %
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