Calculate ppm of NH4+ in (NH4)2SO4 Solution

[elemis]

Homework Statement

A solution is prepared by dissolving 0.475g of (NH4)2SO4 in enough of distilled water and the solution is made up to 2.0 dm3

Calculate the ppm of

NH₄⁺

Homework Equations

The Attempt at a Solution

(NH4)₂SO₄ -----> 2NH₄+ + SO₄^2-

0.475/132 = 3.598 *10^-3 moles of Ammonium Sulphate

2*[3.598 *10^-3] = 7.197*10^-3 NH4+ ions

Since concentration = no. of moles / volume

c = 7.197*10^-3 / 2 = 3.598*10^-3 moldm-3

Converting moldm-3 to gdm-3 gives : 0.475 gdm-3

Converting gdm-3 to milligrams per dm3

we get 475 mgdm-3 or 475 ppm

Is this correct ?

 

[Borek]

Looks OK.

 

[elemis]

Looks OK.

Are you the same Borek who's on the Chemicalforums ?

 

[elemis]

Looks OK.

Is that an assertive yes or a subtle message that I have made a mistake.

 

[Borek]

Yes and yes

 

[elemis]

Yes and yes

I knew I'd seen you somewhere else.

Yes my calculations are right or yes it was a subtle message that I made a mistake ?

 

[Borek]

Yes.

Spoiler

Your result is OK.