Solute Particle Count in KBr, HCl &Na_2CO_3

[recon]

Question:

Which of the following aqueous solutions contain(s) the same number of particles of solute as are contained in 250 mL of 2.0 mol/L KBr

(i) 1.0 L of 1.0 mol/L ethanol, $$C_2H_5OH$$
(ii) 250 mL of 3.0 mol/L calcium chloride, $$CaCl_2$$
(iii) 500 mL of 1.0 mol/L hydrochloric acid, $$HCl$$
(iv) 500 mL of 1.0 mol/L sodium carbonate, $$Na_2CO_3$$

The way I see it, there are 0.25 x 2 = 0.5 moles of KBr
and 0.5 x 1 = 0.5 moles of $$HCl$$ and $$Na_2CO_3$$ in (iii) and (iv). Since they all have the same number of moles, they should have the same number of atoms? Is that what they mean by "same number of particles of solute"?

But according to the answer guide, it says that the answers are (i) and (iii).

 

[loseyourname]

$KBr$ will dissociate into $K^+$ and $Br^-$, so there will be 0.5 + 0.5 = 1.0 mol of ions. $C_{2}H_{5}OH$ is not soluble in water, leaving 1.0 x 1.0 = 1.0 mol of molecules. $HCl$ will dissociate into $H^+$ and $Cl^-$, so there will be 0.5 + 0.5 = 1.0 mol of ions.

 

[recon]

Thanks a lot! I understand better now. However, there is still another problem I am stuck with and can't seem to figure out:

A solution contains potassium nitrate and potassium sulphate. The concentration of potassium ions is 0.650 mol/L and of nitrate ions is 0.400 mol/L. What is the concentration of the sulfate ions?

I don't even know where to start? :( Can someone give me hints?

 

[loseyourname]

A solution contains potassium nitrate and potassium sulphate. The concentration of potassium ions is 0.650 mol/L and of nitrate ions is 0.400 mol/L. What is the concentration of the sulfate ions?

You can start by comparing the concentration of sulfate to the concentration of potassium ions (1:2 - sulfate has charge of -2, whereas potassium has a charge of +1). I don't see how you can solve this without knowing the ratio of potassium nitrate to potassium sulfate, though - or, more accurately, the ratio of nitrate to sulfate. Are you sure there is no more information given?

 

[recon]

We are allowed to refer to the Periodic Table of Elements if that is what you mean. Other than that, we are not given anything else.

EDIT: We were given four choices for the answer. They are
A. 0.325 mol/L
B. 0.250 mol/L
C. 0.200 mol/L
D. 0.125 mol/L

 

[Gokul43201]

No more info is required...other than the permission to assume complete (100%) dissociation.

Given :

$$[K^+]_{tot} = 0.65M = [K^+]_{nitrate} + [K^+]_{sulfate}$$

$$[NO_3^-] = 0.4M = [K^+]_{nitrate}$$

$$So, ~~ [K^+]_{sulfate} = 0.65M - 0.4M = 0.25M = 2*[SO_4^{2-}]$$

$$Thus, ~~[SO_4^{2-}] = 0.25M/2 = 0.125M$$

Choice D is the correct answer.

 

[loseyourname]

Okay, that's pretty obvious. Guess I need to get to bed earlier.