Calculate the mass of impure KOH needed to make up 1.20 L of 0.60 mol/L KOH(aq)? Assume the impure KOH is 84% KOH by mass and 16% water.

 

m_{impure KOH}=?...........84%

V_KOH=1.20L

C_KOH=0.60 mol/L

H₂O=16%

 

I tried to do this but i don't think this is right, I don't know what should I do first...

n_{impure KOH}=m/M= 84/56.1056 = 1.497 mol

n_H2O= m/M = 16/18 = 0.89 mol

 

 

How to solve this??

I should start by deciding the mass of KOH in 1.2L of 0.6M solution.

 

Then see if you can write an equation connecting this to the mass of KOH in the impure source. Does the inclusion of water make any difference?

Why did they tell you it was 16%?

Edited June 16, 201312 yr by studiot

we dnt do home work here...but if i will help, try to rearrange the values and write the equation..it works with the KOH mass

Don't worry so much about the water component of the KOH. It's really only the % by mass of KOH that you need to worry about.

 

If it helps, another way of expressing the amount of KOH you have is by saying that in every gram of your impure sample, you have 0.84 grams of KOH. If you know the mass required to make 1.2 L of 0.60 M KOH, then knowing how much of the impure stuff you need is just a simple conversion. For instance, if say I need 0.84 g of KOH to make up my solution, then I would need 1 g of the impure sample. If I need 1.68 g of KOH, I would use 2 g of the impure sample (1.68 / 0.84 = 2).

 

Your first step is, as studiot said, to work out the mass of KOH required to make your solution.