Question

2. While performing an experiment to determine the percentage of water absorbed by raisin
the following data was obtained:
Mass of water taken in the beaker = 50 g
Mass of dry raisins = 5 g
Mass of soaked raisins in water = 8 g
Calculate the percentage of water absorbed by the raisins.

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Answer 1

That is a problem on endosmosis.

Let W1 be the weight of raisins before immersing them in water.

W1=4.0 g

Let W2 be the weight of raisins after immersing in water for few hours.

W2=7.0 g

Therefore, the weight of water absorbed by raisins is W2-W.

W2-W1=3.0 g

The percentage of water absorbed by raisins = [(W2-W1)/W1]*100%

[(W2-W1)/W1]*100% =75 %

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Difference in Mass of raisin = Mass of water gained by the raisin

Change in Mass = 7g - 4g = 3g

Now,

[ (Change in Mass) \ (Initial mass) ]100 = Percent of weight (of the raisin) gained

[ (3g) \ (4g) ]100 = 75%

That's the answer to the percentage weight of the raisin that was absorbed.

I'm not sure what “percentage of water absorbed by the raisins” is exactly. That depends on the amount of water present in the vessel ( I guess ).

If my assumption is right, then the following makes sense; if not, I'm sorry.

Say, “x” g of water is present in the vessel.

Then Percentage of water of vessel absorbed by the raisin is

[ (Mass absorbed by the raisin) \ (Total mass of water) ]100

[ (3g) \ (“x”g) ]100

This answer was not meant to be this long to be honest. Sorry about that.

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