Asample of seaweed contains 2 liters of water and has 100 grams of salt dissolved in its cells. the seaweed is placed in a bucket with 4 liters of water and 300 grams of dissolved salt. what will occur when the system reaches equilibrium?

The seaweed loses 0.5 L of water and undergoes plasmolysis.  

Step-by-step explanation:

In seaweed: [NaCl] = 100 g/2 L = 50 g/l

In bucket:    [NaCl] = 300 g/4 L = 75 g/L

The salt concentration is higher in the bucket.

The water in the bucket is hypertonic compared to that in the seaweed cells.

Water will pass out of the seaweed by osmosis through its semipermeable membranes until the salt concentrations in the seaweed and the bucket are equal.

Let x be the volume that moves out of the seaweed. At equilibrium,

          [NaCl] = [NaCl]

In seaweed = In water

  00/(2 – x)  = 300/(4 + x)      Multiply each side by (2 – x)(4 – x)

   100(4 + x) = 300(2 – x)       Remove parentheses

400 + 100x = 600 – 300x    Add 300x to each side

400 + 400x = 600                 Subtract 400 from each side

         400x = 200                  Divide each side by 700

               x = 0.5

The seaweed will lose 0.5 L of water.

The cells of the seaweed will shrink in a process called plasmolysis.

law of conservation of matter

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)