Concentration and Molarity

Question

In the laboratory, a student adds 12.9 g of chromium(II) nitrate to a 250. mL volumetric flask and adds water to the mark on the neck of the flask. Calculate the concentration (in mol/L) of chromium(II) nitrate, the chromium(II) ion and the nitrate ion in the solution.

[Cr(NO3)2]
[Cr2+]
[NO3-]

Answer 1

In solution, Cr(NO₃)₂ ==> Cr²⁺(aq) + 2NO₃^-(aq)

Molar mass Cr(NO₃)₂ = 176. g / mol

mol Cr(NO₃)₂ = 12.9 g x 1 mol / 176 g = 0.0733 mol

[Cr(NO₃)₂] = 0.073 mol / 250 ml x 1000 ml / L = 0.293 mol/L

[Cr²⁺] = 0.293 mol / L since there is 1 mol Cr²⁺ per mol Cr(NO₃)₂

[NO₃^-] = 2 x 0.293 mol/L = 0.586 mol / L since there are 2 mols of NO₃^- per mol Cr(NO₃)₂