When 50.0 ml of a 0.3000 m agno3 solution is added to 50.0 ml of a solution of mgcl2, an agcl precipitate forms immediately. the precipitate is then filtered from the solution, dried, and weighed. if the recovered agcl is found to have a mass of 0.1183 g, what as the concentration of magnesium ions in the original mgcl2 solution?

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Kailee

Chemistry

30 January, 17:38

When 50.0 ml of a 0.3000 m agno3 solution is added to 50.0 ml of a solution of mgcl2, an agcl precipitate forms immediately. the precipitate is then filtered from the solution, dried, and weighed. if the recovered agcl is found to have a mass of 0.1183 g, what as the concentration of magnesium ions in the original mgcl2 solution?

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Adrianna Le · Answer 1

0.0165 m The balanced equation for the reaction is AgNO3 + MgCl2 = = > AgCl + Mg (NO3) 2 So it's obvious that for each Mg ion, you'll get 1 AgCl molecule as a product. Now calculate the molar mass of AgCl, starting with looking up the atomic weights. Atomic weight silver = 107.8682 Atomic weight chlorine = 35.453 Molar mass AgCl = 107.8682 + 35.453 = 143.3212 g/mol Now how many moles were produced? 0.1183 g / 143.3212 g/mol = 0.000825419 mol So we had 0.000825419 moles of MgCl2 in the sample of 50.0 ml. Since concentration is defined as moles per liter, do the division. 0.000825419 / 0.0500 = 0.016508374 mol/L = 0.016508374 m Rounding to 3 significant figures gives 0.0165 m