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.

Molecular weight of [Ni(dmg)2] = 232.77

 

Concentration = 0.0273 moles per litre.

I don't have those chemicals/elements memorized anymore, nor do I have a table of them.

You mean a periodic table? Not that hard to come by surely!

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