Question

If I have 0.393g and 0.0393g of CuSO4*5H2O, and separatley dissolve each into 25mL, then dilute 10.00mL of each to a final volume of 100mL, what is the concentration of Cu2+ per 250mL in each solution?

Answer 1

Molar mass of CuSO4.5H2O = 250 g/mol

for 0.393 g CuSO4.5H2O :
number of moles = 0.393/250 = 1.572*10^-3 mol
when dissolved in 25 ml,
C1 = number of moles / volume in L
      = (1.572*10^-3 mol) / 0.025 L
      = 0.063 M

when 10 mL of this is diluted to 100 mL, concentration be C2
use:
C1*v1 = C2*v2
0.063*10 = C2 * 100
C2= 6.3*10^-3 M

so, concentration is 6.3*10^-3 mol/L or 6.3*10^-3 mol / 1000 mL
=6.3*10^-3/4
= 1.6*10^-3 mol/250 mL

for 0.0393 g CuSO4.5H2O :
number of moles = 0.0393/250 = 1.572*10^-4 mol
when dissolved in 25 ml,
C1 = number of moles / volume in L
      = (1.572*10^-4 mol) / 0.025 L
      = 0.0063 M

when 10 mL of this is diluted to 100 mL, concentration be C2
use:
C1*v1 = C2*v2
0.0063*10 = C2 * 100
C2= 6.3*10^-4 M

so, concentration is 6.3*10^-4 mol/L or 6.3*10^-4 mol / 1000 mL
=6.3*10^-4/4
= 1.6*10^-4 mol/250 mL