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A student recorded the mass of dry raisins as 3.0 g and the mass of raisins after soaking in water is 4.8 g. Find the percentage of water absorbed by

raisins during the experiment.



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


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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