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

2 Answers
Jan 17, 2018

Answer:

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.

Jan 17, 2018

Answer:

#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#