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A sample of seaweed contains 1 liter of water and has 50 grams of salt dissolved in its cells. The seaweed is placed in a solution made of 2 liters of water and 150 grams of dissolved salt. What will occur when the system reaches equilibrium?

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seaweed having 50 g salt in 1 L water. The bucket contains 150 g of salt in 2 L of water

amount of water present in bucket is twice to amount of water in weed


V¬_water bucket=2×V_water weed

At equilibrium, volume of water in weed is x and volume in bucket is y but concentration remain same as follows:


50 /x=150 /y</p><p>\\ Y = 3 x

At equilibrium, weed loose z L from 1 L water to bucket containing 2 L as follows:


(2 + z) = 3 (1-z)</p><p>\\ (2 + z) = 3 – 3 z</p><p>\\ Z = 0.25 L

Thus, Weed will loose 0.25 L of water

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