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A student recorded the weight of dry raisins as 5.4g and on soaking them for 4-5 hrs the weight was 7.6g.calculate the percentage of water imbibed by the



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Dry weight of raisins(W_d) = 5.4g
weight weight of raisins(W_w) = 7.6g
weight of water imbibed, w=W_w-W_d=7.6-5.4 = 2.2g

% of water imbibed =  \frac{weight\ of\ water\ imbibed}{dry\ weight\ of\ raisins} \times 100 = \frac{w}{W_d} \times 100= \frac{2.2}{5.4} \times 100=40.74 

Hence, Percentage of water imbibed is 40.74%
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Weight of dry raisins=5.4g
weight of raisins after soaking them=7.6g
% of water imbided by raisins=(weight of wet raisins-weight of dry raisins÷weight of dry raisins)×100
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