# What is the concentration of copper when a 2*g mass of the metal is dissolved in 400*mL of water?

Jul 26, 2017

There are certain assumptions we have to make with regard to your question.

#### Explanation:

Copper is not an active metal, and certainly does not reduce water under standard conditions. To get the $\text{molarity}$, or $\text{molality}$ of a copper salt dissolved in water, say $C u {\left(N {O}_{3}\right)}_{2}$, which is quite water-soluble, we take the quotients....

$\text{Molarity}$ $=$ $\text{Moles of solute (mol)"/"Volume of solution (L)}$

$\text{Molality}$ $=$ $\text{Moles of solute (mol)"/"Kilograms of solvent (kg)}$

If we solve for each quantity, we gets.....according to the boundaries I set.

$\text{Molarity} = \frac{\frac{2 \cdot g}{187.56 \cdot g \cdot m o {l}^{-} 1}}{0.400 \cdot L} = 0.027 \cdot m o l \cdot {L}^{-} 1.$

$\text{Molality} = \frac{\frac{2 \cdot g}{187.56 \cdot g \cdot m o {l}^{-} 1}}{0.400 \cdot k g} = 0.027 \cdot m o l \cdot k {g}^{-} 1.$

And, as in all DILUTE solutions, $\text{molarity}$, and $\text{molality}$ are the same to a first approx.

The density of each solution, would be a LITTLE over $1 \cdot g \cdot m {L}^{-} 1$ (because the volume is not assumed to have changed). The densities you quote are quite unreasonable. At any rate, if you want, you should resubmit this question, with appropriate parameters and boundary conditions.