Question

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

Answer 1

Approx. `26*g` of salt.....

Explanation:

We use the defining quotient....

`"concentration"="moles of solute"/"volume of solution"`

And thus `"moles of solute"="volume"xx"concentraion"`

And for `"moles of solute"`, we gots.....

`652*cancel(mL)xx10^-3*cancel(L*mL^-1)xx0.250*mol*cancel(L^-1)=`

`0.163*mol`...

And with respect to `"copper sulfate"` there is a mass of `0.163*molxx159.61*g*mol^-1=26.0*g`...

Of course in an aqueous solution of copper sulfate, the copper is present as `[Cu(OH_2)_6]^(2+)`, a beautifully blue-coloured complex ion.

Answer 2

`m=ul ?g CuSO_4`

Explanation:

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

`M=eta/V`

where:

`eta=("mass"(m))/("molar mass"(Mm))`

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=((m)/(Mm))/V`; cross multiply to simplify

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

`m=MVMm`; now, plug in the values of each variable

where:

`M=0.250M=(0.250mol)/(L)`

`V=652cancel(mL)xx(1L)/(1000cancel(mL))=0.652L`

`Mm_(CuSO_4)=(159.5g)/(mol)`, obtainable from the periodic table.

`m=(0.250cancel(mol))/cancel(L)xx0.652cancel(L)xx(159.5g)/cancel(mol)`

`m=ul ?g CuSO_4`