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Quantitative: (NH4)2Ni(SO4)2•6H2O - calculate NH4+?

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Hello, I've never been good in quantitative chemistry and once again this kind of question seems to be a trouble for me ehh sad.gif

I would be so grateful for showing the exact working.

0.040 mol of (NH4)2Ni(SO4)2•6H2O is dissolved in water to give 200 cm3 of aqueous solution.
What is the concentration, in mol dm–3, of ammonium ions?

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Hello, I've never been good in quantitative chemistry and once again this kind of question seems to be a trouble for me ehh sad.gif

I would be so grateful for showing the exact working.

0.040 mol of (NH4)2Ni(SO4)2•6H2O is dissolved in water to give 200 cm3 of aqueous solution.

What is the concentration, in mol dm–3, of ammonium ions?

Okay so you would use the formula C (concentration) = n (number of moles) / V (Volume in dm3)

So first you would convert 200 cm3 to dm3 which would be 0.2 dm3

annd then simply substitute the number you have in the formula :) so C = 0.04/0.2 = 0.2 moles.

However, you have 2 of the NH4 so you multiply 0.2 x 2 = 0.4. There's your answer! :D 

Hope this helps!

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