# How many grams of CuSO_4 are present in 652 mL of .250 M CuSO_4 solution?

Jan 17, 2018

Approx. $26 \cdot g$ of salt.....

#### Explanation:

We use the defining quotient....

$\text{concentration"="moles of solute"/"volume of solution}$

And thus $\text{moles of solute"="volume"xx"concentraion}$

And for $\text{moles of solute}$, we gots.....

$652 \cdot \cancel{m L} \times {10}^{-} 3 \cdot \cancel{L \cdot m {L}^{-} 1} \times 0.250 \cdot m o l \cdot \cancel{{L}^{-} 1} =$

$0.163 \cdot m o l$...

And with respect to $\text{copper sulfate}$ there is a mass of $0.163 \cdot m o l \times 159.61 \cdot g \cdot m o {l}^{-} 1 = 26.0 \cdot g$...

Of course in an aqueous solution of copper sulfate, the copper is present as ${\left[C u {\left(O {H}_{2}\right)}_{6}\right]}^{2 +}$, a beautifully blue-coloured complex ion.

Jan 17, 2018

m=ul ?g CuSO_4

#### Explanation:

Given the molarity$\left(M\right)$ and the volume$\left(V\right)$ of the solution, the number of moles$\left(\eta\right)$ can be solved using the formula shown below:

$M = \frac{\eta}{V}$

where:

$\eta = \left(\text{mass"(m))/("molar mass} \left(M m\right)\right)$

Now, plug in this value to the formula. Rearrange formula to isolate $\eta$. Always attached unit to each variable's value to ensure that the desired unit is correctly labeled to the required variable; that is,

$M = \frac{\frac{m}{M m}}{V}$; cross multiply to simplify

MV=m/(Mm; isolate the required variable $m$

$m = M V M m$; now, plug in the values of each variable

where:

$M = 0.250 M = \frac{0.250 m o l}{L}$

$V = 652 \cancel{m L} \times \frac{1 L}{1000 \cancel{m L}} = 0.652 L$

$M {m}_{C u S {O}_{4}} = \frac{159.5 g}{m o l}$, obtainable from the periodic table.

$m = \frac{0.250 \cancel{m o l}}{\cancel{L}} \times 0.652 \cancel{L} \times \frac{159.5 g}{\cancel{m o l}}$

m=ul ?g CuSO_4