Question

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.

Answer 1

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

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

60% water was absorbed.

Answer 2

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.