The following data was obtained on performing an experiment for determining the percentage of water absorbed by raisins:

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

The percentage of water absorbed by raisins will be:

(1) 10%
(2) 25%
(3) 45%
(4) 50%

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Solution

The precentage of water absorbed is calculated as -

$\frac{Final weight - Initial weight}{Initial weight} \times 100$

$\frac{30 - 20}{20} \times 100 = \frac{1000}{20} = 50 %$

Hence, the correct answer is option 4.

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Similar questions

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:

Dry raisins were soaked in water for 2 hours, to determine the percentage of water absorbed by raisins. Before final weighing of swollen raisins, the extra water left on the surface of soaked raisins was removed by:

Q. A student soaked 5 g of raisins in beaker (A) containing 25 mL of ice-chilled water and another 5 g of raisins in beaker (B) containing 25 mL of tap water at room temperature. After one hour the student observed that:
(1) water absorbed by raisins in beaker (A) was more than that absorbed by raisins of beaker (B).
(2) water absorbed by raisins in beaker (B) was more than that absorbed by raisins of beaker (A).
(3) the amount of water absorbed by the raisins of both beakers (A) and (B) was equal.
(4) no water was absorbed by raisins in either of the beakers (A) and (B).

Q. 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.

Q. Each of the three beakers A, B and C contained 50 mL of distilled water. A student placed five raisins in each beaker. The raisins for each beaker weighed the same. The beakers were kept at room temperature. The raisins were removed from beaker A after 10 minutes, from beaker B after 20 minutes and from beaker C after one hour. On calculating the percentage of water absorbed by the raisins, it was found that:
(1) maximum absorption of water by raisins was in beaker C
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(4) absorption of water was equal in raisins of all the three beakers

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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

The precentage of water absorbed is calculated as - Final weight-Initial weightInitial weight×100 30 -2020 ×100 = 100020 = 50% Hence, the correct answer is option 4.

The precentage of water absorbed is calculated as -

$\frac{Final weight - Initial weight}{Initial weight} \times 100$

$\frac{30 - 20}{20} \times 100 = \frac{1000}{20} = 50 %$

Hence, the correct answer is option 4.