Question

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 raisins

Answer 1

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%

Answer 2

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