Molarity problem?

Question

I'm having a tough time with all of it but this I just haven't been able to figure out. This isn't homework just prep for a test.

A reagent bottle is labeled 0.450 M K(2)CO(3)

c) Assuming no volume change, how many grams of K2CO3 do you need to add to 2.00 L of this solution to obtain 1.000 M solution of K2CO3?

The answer in the back of the book is 152 g K2CO3. I've been trying for the life of me to get this.

Answer 1

Atomic weights: K=39.1 C=12 O=16 K2CO3= 138.2
Let the original solution be called S; let K2CO3 be called K

2.00LS x 0.450molK/1.00LS = 0.900molK

So in the 2L of S, we have 0.900molK. But to make 2L of 1.000MK we need 2.000molK. So we need 1.100molK more.

1.100molK x 138.2gK/1molK = 152gK

Answer 2

Ok....managed ot get your answer and here's how

To start with you need the molecular weight (Mr) of K2CO3 which is:

(2 x K) + (1 x C) + (3 x O) = (2 x 39) + (1 x 12) + (3 x 16)
= 138
The Mr is equivalent to the mass (in g) which would contain 1 mole of your compound.

So..... dissolving 138 g in 1 litre would yield a 1M solution of K2CO3

So, you already have a 0.450M solution

This will contain 0.450 x 138 = 62.1 g of K2CO3 in 1l

You multiply the Mr by 0.45 as this is essentially what you ahve done to a 1M solution to get a 0.45M solution

So, in order to have a 1M solution, you would need to make up the mass of K2C03 to 138 g in 1l. as you are not changing the volume, you must be adding solid compound and so it's a simple subtraction to work it out.

138 - 62.1 = 75.9 g

So you need to add 75.9 g of K2CO3 to 1l of a 0.450M solution to make it up to a 1M solution

Because you have 2 l of this solution, you simply multiply by 2

75.9 x 2 = 151.8 g = 152 g ( to nearest g)

Hope that helps!!

Answer 3

You are starting with .450 moles per liter. Your 2 liters contain 2 x .450 moles or .900 moles of K2CO3. You need an additional 1.10 moles. 1.10 times the molar mass of K2CO3 (138.2 g/mol) is 152 g.