Molarities: 0.345 M, 0.365 M, 0.401 M, 0.389 M Determine the molarity of an ion

Question

Molarities: 0.345 M, 0.365 M, 0.401 M, 0.389 M Determine the molarity of an ion in solution after the addition of an ion that precipitates it given its initial volume and the mass of the dried precipitate Example: 10. You have 100 mL of an strontium solution of unknown concentration. You add an excess of a Na3PO4 solution unitl all of the strontium forms a precipitate in the form of Sr3(PO4)2. You collect and dry the precipitate and determine the mass to 4.56 g. What was the concentration of the strontium ion in the unknowrn solution. alculate the wavelengths of light needed for transitions of electrons in drogen atom. You will be given the constants and the equation need ample 11. Calculate the wavelength of light needed to promote an electror from n = 1 to n = 3 in the hydrogen atom.

Explanation / Answer

Molecular weight of Sr3(PO4)2 = 452.8 gm/mol

atomic weight of Sr = 87.62 gm/mol

mass of the precipitate obtained = 4.56 gm

volume of solution = 100 ml = 100/1000 l = 0.1 L

So,

in 452.8 gm of Sr3(PO4)2, Strontium present is = 87.62 x 3 = 262.86 gm

in 1 gm of Sr3(PO4)2, Strontium present is =  262.86 / 452.8 = 0.5805 gm

so in 4.56 gm of Sr3(PO4)2, Strontium present is = 0.5805 x 4.56 = 2.64 gm

Therefore 2.64 gm of strontium ion was present in the solution.

hence molarity of given strontium ion solution is = mass of strontium in solution/ atomic mass of strontium x Volume of solution in litre = 2.64/ 87.62 X 0.1 = 0.30 M

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