- #1

- Homework Statement
- You have bottle with an unknown solution, you take 0.801g from it and put in a beaker, you add 43.883g of distilled water to the beaker. If you take 2.34g from the beaker, how many grams would that be if you had taken from the bottle instead with the unknown solution.

- Relevant Equations
- C1V1=C2V2

Hi,

So I am aware of the C1V1=C2V2 equation but I feel that this is not really applicable here.

I attempted to go about it through instinct, so I would take 0.801g/43.883g = 0.0182. And then just multiply 0.0182 with 2.34g, to get 0.0426g.

So the answer would be, if I took 2.34 g from the beaker, I could have instead taken 0.0426g from the bottle with the unknown solution.

No idea if that is correct, what do you think?

But what also is bothering me is, if someone ask me, why not take 0.801/(43.883+0.801) as the ratio instead? I wouldn't know how to answer that..

Is there an equation I can refer to?

Thanks for your time

So I am aware of the C1V1=C2V2 equation but I feel that this is not really applicable here.

I attempted to go about it through instinct, so I would take 0.801g/43.883g = 0.0182. And then just multiply 0.0182 with 2.34g, to get 0.0426g.

So the answer would be, if I took 2.34 g from the beaker, I could have instead taken 0.0426g from the bottle with the unknown solution.

No idea if that is correct, what do you think?

But what also is bothering me is, if someone ask me, why not take 0.801/(43.883+0.801) as the ratio instead? I wouldn't know how to answer that..

Is there an equation I can refer to?

Thanks for your time