5. While performing an experiment to determine the percentage of water absorbed by raisins, [0. the following data was obtained: Mass of water taken in the beaker \( =50 g \) Mass of raisins before soaking them in water \( =5 g \) Mass of raisins after soaking in water for 2 hours \( =8 g \). The percentage of water absorbed by raisins would be : a) \( \frac{(8-5) g}{8 g} \times 100 \) b) \( \frac{(8-5) g}{5 g} \times 100 \) c) \( \frac{(8-3) g}{8 g} \times 100 \
5. While performing an experiment to determine the percentage of water absorbed by raisins, [0. the following data was obtained: Mass of water taken in the beaker \( =50 g \) Mass of raisins before soaking them in water \( =5 g \) Mass of raisins after soaking in water for 2 hours \( =8 g \). The percentage of water absorbed by raisins would be : a) \( \frac{(8-5) g}{8 g} \times 100 \) b) \( \frac{(8-5) g}{5 g} \times 100 \) c) \( \frac{(8-3) g}{8 g} \times 100 \) d) \( \frac{(8-5) g}{3 g} \times 100 \)
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The answer is (b)
The percentage of water absorbed by raisins is calculated by using the formula
W2-W1/W1 x 100
where W2 is the weight of wet raisin and W1 is the weight of dry raisin.
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