Chemistry
posted by Anonymous .
HNO3 used as a reagent has a specific gravity of 1.42 g/ mL and contains 70% by strength HNO3. Calculate:
Volume of acid that contains 60 g pure acid?
Volume of water required to make 1 molar solution from 2 mL concentrated HNO3?

60g/1.42g/ml= 42.25ml
Molarity concentrated sol= .060/63*1/42.25= 2.25
so you want to dilute it 2.25/1 times, which means one part concentrated HNO#, 1.25 part water.
One part is 2ml, so 1.25 part water is 2.50ml 
I have interpreted the problem differently from Bob Pursley so my answers differ.
First, the pure acid is what molarity?
1.42 x 1000 x 0.70 x (1/63) = 15.8M
Then use c1v1 = c2v2
15.8M x 2 mL = 1M x v2
v2 = about 15.8 x 2/1 = about 31.6 mL = total volume v2 which means you add about 31.62 = about 29.6 mL H2O
Then to obtain 60 g pure HNO3 you will want 1.42 g/mL x ?mL x 0.70 = 60 and solve for ? mL. I obtained something like 60g of the pure acid. 
60g/1.42g/ml= 42.25ml
Molarity concentrated sol= .060/63*1/42.25= 2.25
so you want to dilute it 2.25/1 times, which means one part concentrated HNO#, 1.25 part water.
One part is 2ml, so 1.25 part water is 2.50ml
First, the pure acid is what molarity?
1.42 x 1000 x 0.70 x (1/63) = 15.8M
Then use c1v1 = c2v2
15.8M x 2 mL = 1M x v2
v2 = about 15.8 x 2/1 = about 31.6 mL = total volume v2 which means you add about 31.62 = about 29.6 mL H2O
Then to obtain 60 g pure HNO3 you will want 1.42 g/mL x ?mL x 0.70 = 60 and solve for ? mL. I obtained something like 60g of the pure acid.
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