Answer
The $Cu(NO_3)_2$ final molarity is equal to $ 0.958$ $M$.
Work Step by Step
$50$ mL = $50 \times 10^{-3}$ L = 0.05 L
1. Calculate the molar mass $(Cu)$:
63.55* 1 = 63.55g/mol
2. Calculate the number of moles $(Cu)$
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 3.045}{ 63.55}$
$n(moles) = 0.0479$
3. Use the balance coefficients to find the amount of $Cu(NO_3)_2$:
$0.0479 mol(Cu) \times \frac{3mol(Cu(NO_3)_2)}{3mol(Cu)} = 0.0479 mol (Cu(NO_3)_2)$
4. Find the concentration in mol/L $(Cu(NO_3)_2)$:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 0.0479}{ 0.05} $
$C(mol/L) = 0.958\ M$
