Question

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?

Answer 1

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

\\ 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)

\\ (2 + z) = 3 – 3 z

\\ Z = 0.25 L$

Thus, Weed will loose 0.25 L of water