0.8 gram of divalent metal was dissolved in 100 CC of 1.28 N HCl and the solution was diluted to to 200 cc. then 50cc of this solution required 54.6 CC of 0.22N NaOH for the neutralization. find the atomic weight of metal .

The excess of HCl is titrated with NaOH. Thus, we can solve the moles of HCl that reacts with the metal and the moles of the metal. As the atomic weight is the ratio between the mass and moles of the atom:

Moles NaOH:

0.0546L * (0.22mol / L) = 0.0120 moles NaOH = Moles HCl in excess

For NaOH and HCl solutions, Normality is the same than molarity.

Moles HCl added:

0.100L * (1.28mol / L) = 0.128 moles HCl * (50mL / 200mL) = 0.032 moles

Moles HCl in that react:

0.032 moles - 0.012 moles = 0.020 moles HCl

Moles M:

0.020 moles HCl * (1mol M / 2 mol HCl) = 0.010 moles M

Molar mass M:

0.8g / 0.010mol =

0.8 gram of divalent metal was dissolved in 100 CC of 1.28 N HCl and the solution was diluted to to 200 cc. then 50cc of this solution required 54.6 CC of 0.22N NaOH for the neutralization. find the atomic weight of metal .​

Respuesta :

80g/mol is the molar mass of the metal

Otras preguntas

0.8 gram of divalent metal was dissolved in 100 CC of 1.28 N HCl and the solution was diluted to to 200 cc. then 50cc of this solution required 54.6 CC of 0.22N NaOH for the neutralization. find the atomic weight of metal .