a student recorded the mass of dry raisins is 6.0 g and mass of raisins after soaking them in water in 10.5 g. calculate the percentage of water absorbed by raisins.

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Solution

Initial weight , x = 6g
Final weight , y = 10.5 g
Mass percentage of water absorbed = $\frac{F i n a l w e i g h t - I n i t i a l w e i g h t}{I n i t i a l w e i g h t} \times 100$
Mass percentage of water absorbed = y-x/x×100
=10.5 -6 /6 ×100 =4.5/6 ×100 =3/4 ×100 =75 percent .
Raisins swell because up because they have semi permeable membrane and they take up water due to osmosis. Hope this helps

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While performing the experiment with raisins to determine the percentage of water absorbed by them, a student made following measurements:

Mass of water in the beaker = 40 g

Mass of raisins before soaking = 5 g

Mass of raisins after soaking for 2 hours = 8 g

Mass of water left in the beaker after the experiment = 35 g

The percentage of water absorbed by raisins is:

Mass of water in the beaker	=	50 g
Mass of dry raisins	=	20 g
Mass of raisins after soaking in water	=	30 g
Mass of water left in the beaker after the experiment	=	40 g

Ram soaked 8g of raisins in water and after 4 hours found that their mass has been 12g . Determine the percentage of water absorbed by raisins.

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