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2015-03-11T18:24:50+05:30

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Mass of dry raisins, m_d= 3.0g
mass of wet raisins, m_a= 4.8g
mass of water absorbed, wm_a-m_d = 4.8-3.0 = 1.8g

% of water absorbed =  \frac{w}{m_d} \times 100= \frac{1.8}{3.0} \times 100=60

60% water was absorbed.
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2015-03-11T18:34:27+05:30
Given,

The mass of dry raisins
: 3.0 gram
The mass of water-soaked raisins : 4.8 grams

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Amount of water, the raisins soaked = (4.8-3.0) grams
                                                     = 1.8 grams

Amount of water, the raisins soaked in percentage = 1.8/3.0 * 100
                                                                                 = 60.0 % (Ans.)
                                                                                  


Answer: Therefore, the raisins soaked 60% of water.




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