Respuesta :
Answer:
concentrate of working solution = 730 × [tex]10^{-3}[/tex] mol/L
Explanation:
given data
zinc nitrate volume = 240 mL = 0.240 L
volume of stock solution = 70 mL = 0.07 L
concentration of stock solution = 2.5 mol/L
to find out
concentration of the chemist's working solution
solution
first we get here mole of zinc nitrate that is
mole of zinc nitrate = volume × concentrate .......1
put here value we get
mole of zinc nitrate = 0.07 × 250
mole of zinc nitrate = 0.175 moles
now we get concentrate of working solution that is
concentrate of working solution = [tex]\frac{mole\ of\ zinc\ nitrate}{zinc\ nitrate\ volume}[/tex] ..............2
put here value we get
concentrate of working solution = [tex]\frac{0.175}{0.240}[/tex]
concentrate of working solution = 730 × [tex]10^{-3}[/tex] mol/L
Answer:
M = 0.729 M
Explanation:
In this case, we have a solution of Zn(NO₃)₂.
The chemist wants to prepare a dillute solution of this reactant.
The stock solution of the nitrate has a concentration of 2.5 M, and he wants to prepare 240 mL of a more dillute concentration of the same solution. He adds 70 mL of the stock and complete with water until it reach 240 mL.
We want to know the concentration of this dilluted solution.
As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the dilluted solution so:
n1 = n2 (1)
and we also know that:
n = M*V (2)
If we replace this expression in (1) we have:
M1*V1 = M2*V2
Where 1, would be the stock solution and 2, the solution we want to prepare.
So, we already know concentration and volume used of the stock solution and the desired volume of the dilluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is M2:
2.5 * 70 = 240M2
M2 = 2.5 * 70 / 240
M2 = 0.729 M
This is the concentration of the solution prepared.
