wolfson Posted December 15, 2003 Share Posted December 15, 2003 Thought i would do a chem quiz (feeling nerdy): 50.00 mL of a solution of nickel bromide was treated with an excess of dimethylglyoxime solution. The precipitated Ni(DMG)2 was filtered off; 0.3177g was obtained. Calculate the concentration (molarity), of the nickel bromide solution. I will say when i see the correct answer. Link to comment Share on other sites More sharing options...
greg1917 Posted December 15, 2003 Share Posted December 15, 2003 Molecular weight of [Ni(dmg)2] = 232.77 Concentration = 0.0273 moles per litre. Link to comment Share on other sites More sharing options...
Dudde Posted December 15, 2003 Share Posted December 15, 2003 I don't have those chemicals/elements memorized anymore, nor do I have a table of them. Link to comment Share on other sites More sharing options...
greg1917 Posted December 15, 2003 Share Posted December 15, 2003 You mean a periodic table? Not that hard to come by surely! Link to comment Share on other sites More sharing options...
wolfson Posted December 16, 2003 Author Share Posted December 16, 2003 Mr Ni(C4H7N2O2)2 = 288.93 g mol-1 Amount Ni(dmg)2 = mass/molar mass = 0.3177 g/288.93 g mol-1 =1.100 x 10-3 mol Ni(dmg)2 1 mol NiBr2 gives 1 mol Ni(dmg)2 so amount NiBr2 = 1.100 x 10-3 mol Concentration of NiBr2 = amount/volume = 1.100 x 10-3 mol/50.00 x 10-3 L = 0.02199 M NB The final answer to 4 sf is 0.02199 M not 0.02200 M, because although I have written the intermediate result as 1.100 x 10-3 mol, I have not actually rounded it to 4sf at this point. The figure on the calculator is 1.09957 x 10-3, and I have used this figure in the subsequent calculation and rounded off only at the last step. In this case it doesn't make a lot of difference, but in highly precise analytical work it may do. very close greg Link to comment Share on other sites More sharing options...
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