Posted by smiley on Friday, August 31, 2007 at 12:24pm.
Orgwt*percentsample-(.5117-.4998)= water left, and water left= .1*.4998
or
percentsample= (.0119+.1*.4998)/.5117
and if do not want decimal percent, then muliply by 100
So, the moisture content (as received) is ((0.5117-0.4998)/0.5117)* 100 = 2.326%
and the the moisture content (dry basis) is ((0.1*0.4998)/0.5117)*100 = 9.767%
and when the sample is completely dried, it would weigh 0.4998-(0.1*0.4998) = 0.4498
Am I right?
I am having a problem understanding this problem.
A 0.5117 gram fertilizer containing phsphorus was oven dried. The weight of the dried sample is 0.4998 grams. However, 10% moisture is still retained in the sample. Calculate the moisture content in the sample in as received and dry basis. If the sample is completely dried, how much wouldit weigh?
Is the 10% moisture retained after partial drying 10% of 0.4498 OR 10% of 0.5117.
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