Question

What are the respective concentrations (M) of Cu2+ and Cl- afforded by dissolving 2.0 g of CuCl2 in water and diluting to 500 mL.

Answer 1

Answer:

$M_{Cu^{2+}}=0.030M \\\\M_{Cl^-}=0.060M$

Explanation:

Hello.

In this case, the first step is to compute the moles of copper (II) chloride (molar mass: 134.45 g/mol) in 2.0 g as follows:

$n_{ClCl_2}=2.0CuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2} =0.015molCuCl_2$

Thus, since one mole of copper (II) chloride contains 1 mole of copper (its subscript) and 2 moles of chloride (its subscript), those moles are respectively:

$n_{Cu^{2+}}=0.015molCuCl_2*\frac{1molCuCl_2}{1molCuCl_2} =0.015molCu^{2+}\\\\n_{Cl^-}=0.015molCuCl_2*\frac{2molCl^-}{1molCuCl_2} =0.030molCl^-$

Therefore, the concentrations (in molar units) considering the volume in liters of the solution (0.500 L for 500 mL) are:

$M_{Cu^{2+}}=\frac{0.015molCu^{2+}}{0.500L}=0.030M \\\\M_{Cl^-}=\frac{0.030molCu^{2+}}{0.500L}=0.060M$

Best regards.