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Hi everyone,

I was just wondering if you can answer this chemistry question for me?

I need to make 2M Ammonium Molybdate and only 250mls of it (for a senior chem class).

If the Molecular weight on the bottle is 1235.86, I know i have to times that by 2 for 2M and then divide total by 4 for a 250ml solution.

The answer is 617.93g of powder. Ok...that is a lot of powder.

How much water do i add...... 250mls???? Then its not going to be a 250ml solution is it? Its going to be around 800ml i am guessing because of the huge amount of powder.

But its supposed to be 250mls?

Any thoughts.....my teachers are all stumped.

2M Ammonium Molybdate is impossible to make.
The solubility of Ammonium Molybdate is approximately 50gm/100mls
125 gms at the most will dissolve in 250mls.
What does she need it for?
The 0.1 M solution I have in the Chem Store is not all dissolved.
For the record, to make up 250 mls of a solution I would use a 250ml volumetric flask.
Using the amount of chemical required to make 250 mls and then distilled water up to the line.

What are they using it for? We don't ever use above 0.5m

Its for Yr 12 Chemistry - Qualitative tests for assessing water quality. I haven't got any so teacher will have to fudge it !
Not sure if we can use any other "molybdate"
The prac from conquering Chemistry doesn't give a molarity.

We use 0.1M ammonium molybdate for that....works OK.