A0.05270.0527 g piece of metal containing an unknown amount of silver is dissolved to form 25.00 ml of solution. to the solution, 35.435.4 ml of 0.02190.0219 m kcl kcl is added, resulting in the complete precipitation of the silver ash agcl agcl . the excess cl−cl− added to the solution is then back titrated with a standard 0.02190.0219 m agno3 agno3 solution. a total of 14.3314.33 ml of the standard nano3 agno3 solution is required to reach the silver chromate end point in the mohr method. calculate the percent silver in the piece of metal.

93.30%

Explanation:

The equilibrium that takes place is:

First, let's calculate the total moles of Cl⁻ added to the sample:

An unknown amount of this moles reacted with the sample, while the excess was back titrated with a solution containing Ag⁺:

Thus, the moles of Cl⁻ that reacted with the sample are:

Then in the 0.0527 g sample there are 0.461 * 10⁻³ mol Ag, we calculate the mass of Ag in the sample:

Finally we calculate the percent silver:

the grams   of   barium that   are in   2.3 x 10^23   atoms of barium   is   52.4 g (answer c)

    calculation

mass = moles   of ba   x molar   mass of ba

  molar mass of ba from periodic table=   137.3 g/mol

  use the   avogadro's law   constant   to   calculate the moles of barium

  that   is     1 mole =   6.02   x 10^23 atoms

                    ? moles = 2.3 x 10^23   atoms

by cross   multiplication

=(1 mole   x 2.3x 10^23 atoms)   / 6.02   x10^23 atoms)=   0.382   moles

mass   is therefore = 137.3   g/mol x 0.382 mol = 52.4 ( answer c)

explanation: