Calculate Acetic Acid Adsorbed by Activated Charcoal

[Hydrous Caperilla]

Homework Statement

(Q.) 3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is :

Homework Equations

The Attempt at a Solution

I thought of finding the volume first by using the formulae :N1V1=N2V2 and then get the volume but I realized that finding volume won't help so I am out of ideas

 

[Borek]

The volume doesn't change.

How many moles of the acetic acid before adding the charcoal, how many moles after adsorption?

 

[Hydrous Caperilla]

the n factor for acetic acid is 1 so it's molarity will be same (o.o6M). and from the formulae ,we can calulate the moles but I don't know how to calculate moles after adsorption

 

[Borek]

You have 50 mL of a 0.042 M solution, how do you calculate number of moles in such case?

What is the definition of the molar concentration?

 

[Hydrous Caperilla]

You have 50 mL of a 0.042 M solution, how do you calculate number of moles in such case?

Molarity X Volume=0.042 X 0.05

 

[Hydrous Caperilla]

The volume doesn't change.

Doesn't volume increase with addition of solutes

 

[Borek]

Doesn't volume increase with addition of solutes

In general you are right, it does. But in most cases change can be neglected.

Please pay attention to what you post, each your post contains unnecessary multiple quotes (I will remove them now). Use "preview" button to see what it will look like. You can also edit your posts for a short period of time.

 

[Hydrous Caperilla]

In general you are right, it does. But in most cases change can be neglected.

Please pay attention to what you post, each your post contains unnecessary multiple quotes (I will remove them now). Use "preview" button to see what it will look like. You can also edit your posts for a short period of time.

So I want to post the steps I have understood

Step 1 Before Adsorption

No of moles of acetic acid in the solution=50X0.6=3milli moles(n factor of acetic acid is 1 so Normality=Molarity)

Step 2
No of mole of acetic acid after adsorption=50(volume is not changing which I don't understand)X0.042=2.1 millimoles

Step 3
Moles adsorbed= O.9millimoles

Step 4

Weight of acetic acid=0.9X60/1000=54mg
My textbook gives the answer as 18mg...what did I do wrong now and can you tell me more about when to keep volume constant

 

[Borek]

Weight of acetic acid=0.9X60/1000=54mg

That's correct.

My textbook gives the answer as 18mg

Looks OK.

what did I do wrong

Reread the question, you have missed something.

when to keep volume constant

Almost always, only when the concentration changes are huge it really matters.

0.06 M acetic acid contains 3.6 g of acetic acid per liter.
0.042 M acetic acid contains 2.5 g of acetic acid per liter.
The difference is 1.1 g. Imagine removing the acid from the solution - you are removing 1 g from 1000 g, so you can expect volume change in the 0.1% range at best. Negligible.

 

[Hydrous Caperilla]

I get it now.
I forgot to divide it by the mass of charcoal so it makes sense now.
Thanks