Question

What are the respective concentrations (M) of Cu2+ and Cl- afforded by dissolving 0.833 mol CuCl2 in water and diluting to 375 mL?

a. 0.00175 and 0.00175

b. 1.75 and 0.00175

c. 0.570 and 0.198

d. 0.00175 and 1.75

e. 1.75 and 3.51

Answer 1

From one mole of CuCl2 we get one mole Cu2+ and 2 mole Cl-

$\small CuCl_2 \rightarrow Cu^{2+}+2Cl^-$

thus the concentration of Cu2+ will be equal to the concentration of CuCl2 and the concentration of Cl- will be twice the concentration of CuCl2

So the answer of this question will be (e) 1.75 and 3.51

*****I think there is something wrong with the data or the options, because the actual answer should be like below

concentration of the solution = $\small \frac{Number.Of.Moles}{Volume.in.L}$

given the volume = 375 mL = 0.375 L

and the number of moles of CuCl2 = 0.833 moles

so the concentration = $\small \frac{0.833}{0.375}$ = 2.21 M

hence the concentration of Cu2+ = 2.21 M and the concentration of Cl- = 2*2.21M = 4.42 M