# A sample of seaweed contains two 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?

##### 1 Answer

The seaweed must lose

#### Explanation:

This is a great example of how an osmosis problem would look like.

However, you don't *actually* have to use the concept of osmosis to solve this one. Here's why I say that.

So, you know that your seaweed contains

The first thing to notice here is that **initially**, the volume of water in the bucket in **twice as large** as the volume of water in the seaweed.

#V_"water b" = 2 xx V_"water sw"#

Now, when **equilibrium is established**, the volume of water in the seaweed will be *concentration* of salt must be the same for both the bucket and the seaweed!

This means that you can say

#100/x = 300/y implies y = 3x#

So **at equilibrium**, the volume ratio between the volume of water in the bucket and the volume of water in the seaweed has **increased**. This means that the seaweed **lost water** to its surroundings, i.e. to the bucket.

If **at equilibrium**

#overbrace((4 + z))^(color(red)("water in the bucket")) = 3 xx overbrace((2 - z))^(color(blue)("water in the seaweed"))#

Solve this for

#4 + z = 6 - 3z#

#4z = 2 implies z = 2/4 = "0.5 L"#

Therefore, the seaweed **lost**

Now, let's see if our knowledge of how osmosis works can confirm this result.

As you know, **osmosis** is defined as the movement of **solvent** molecules through a semi-permeable membrane from a region of **lower solute concentration** to a region of **higher solute concentration**.

The initial concentrations of salt in the seaweed and in the bucket are

#"100 g"/"2 L" = "50 g/L" -># in theseaweed

#"300 g"/"4 L" = "75 g/L" -> # in thebucket

**SIDE NOTE** *You can convert these concentrations into percent concentrations by mass or into mass by volume percent concentrations if you want, the result will be the same.*

So, the bucket has a **higher** concentration of solute than the seaweed. This tells you that when osmosis kicks in, the seaweed will **lose water**.

From this point on, the calculations are exactly the same. If **lost** by the seaweed, you can say that **at equilibrium**

#100/(2-z) = 300/(4 + z)#

Solve this for

#100 * (4 + z) = 300 * (2- z)#

#400 + 100z = 600 - 300z#

#400z = 200 implies z = 200/400 = 0.5#

The first result is confirmed - in order for equilibrium to be obtained, the seaweed must **lose**