Is that 65% and 17% w/w or w/v?
Do you have a density listed.
1) How do I prepare 2M nitric acid from 65% concentrated nitric acid?
2) How do I prepare 17% ammonia from 25% concentrated ammonia?
Please show calculation steps. Thanks.
6 answers
Assuming it is w/w as is the case for most acids and conc ammonia, can you show me the steps using 69% HNO3 and 28% ammonia?
I have density of 69% HNO3 as 1.42g/ml & 28% ammonia as 0.88g/ml.
I have density of 69% HNO3 as 1.42g/ml & 28% ammonia as 0.88g/ml.
I would start the HNO3 by determining the molarity of the HNO3.
density = 1.42 g/mL x 1000 mL = 1420 grams.
It is 69% by mass; therefore, the amount of HNO3 in that 1420 g is 1420 x 0.69 = 979.8 grams. How many moles is that?
979.8/molar mass = about 15.5 M
Then mL x M = mL x M
mL x 15.5 M = 100 mL x M and I just made up the 100 mL number since you don't have a preference for the amount. Whatever volume you want goes in there.
The NH3 problem is done the same way.
density = 1.42 g/mL x 1000 mL = 1420 grams.
It is 69% by mass; therefore, the amount of HNO3 in that 1420 g is 1420 x 0.69 = 979.8 grams. How many moles is that?
979.8/molar mass = about 15.5 M
Then mL x M = mL x M
mL x 15.5 M = 100 mL x M and I just made up the 100 mL number since you don't have a preference for the amount. Whatever volume you want goes in there.
The NH3 problem is done the same way.
I agree with the procedure.
My nitric (V) acid is 69.5% translating to a mass of 986.9 grams(pure acid).
Molarity of concentrated solution is 15.7M.
To prepare 2M HNO3 from it,measure 127.4cm³ of concentrated solution and top up to 1litre.
My nitric (V) acid is 69.5% translating to a mass of 986.9 grams(pure acid).
Molarity of concentrated solution is 15.7M.
To prepare 2M HNO3 from it,measure 127.4cm³ of concentrated solution and top up to 1litre.
um
why was this made in 2010???
5 answers