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

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

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Answer 1 · 2017-05-11T09:24:50

$"Concentration"="Moles of solute"/"Volume of solution"......=0.40*mol*dm^-3.$

Explanation:

$"Concentration with respect to the nickel salt"$

$=(0.040*mol)/(0.200*dm^3)=0.20*mol*dm^-3$

But given the formulation of the salt, $(NH_4)_2[Ni (SO_4)_2]*6H_2O$ , there are 2 equiv of ammonium per equiv of salt. And thus $[NH_4^+]=0.40*mol*dm^-3$

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

1 Answer

Explanation:

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