Question

A copper cycle experiment takes copper atoms through reactions that produce copper compounds and complexes one after the other, finally producing elemental copper. Copper atoms are conserved throughout the process. Given that a student begins with 9.29 mL of a 0.14 M Cu(NO3)2 solution, how much copper should be isolated at the end of the cycle?

Answer 1

Answer: 0.082 g

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

Molarity = 0.14 M

n= moles of solute  = ?

$V_s$ = volume of solution in ml = 9.29 ml

$0.14=\frac{n\times 1000}{9.29}$

$n=\frac{0.14\times 9.29}{1000}=1.3\times 10^{-3}moles$

Now as 1 mole of $Cu(NO_3)_2$ contains 1 mole of copper

$1.3\times 10^{-3}moles$ moles of $Cu(NO_3)_2$ contains =$\frac{1}{1}\times 1.3\times 10^{-3}=1.3\times 10^{-3}moles$ of copper

Mass of copper =$moles\times {\text {molar mass}}=1.3\times 10^{-3}\times 63.5=0.082g$

Thus 0.082 g of copper should be isolated at the end of the cycle.