posted by Henry .
How would you make up 255 mL of 0.150 M HNO3 from nitric acid that is 68.0% HNO3?
The density of 68.0% HNO3 is 1.41 g/mL.
someone help me on this plox??
Supposing the "68%" is 68% by weight:
(0.255 L) x (0.150 mol/L) x (63.0130 g/mol) / 0.680 / (1.41 g/ml) = 2.51 ml
The direct way would be to take 2.51 ml of the 68% solution and dilute it to exactly 255 mL.
But as a practical matter, since measuring 2.51 mL accurately might be difficult, it would be useful to make an intermediate dilution of something like 10 to 1, and use the diluted solution to make up the 255 ml of desired concentration.
I agree with the answer by james but I would like to point out two or three things. First, the percent IS by weight. Second, the 68% is never EXACTLY 68%. Third, since you can't easily measure out 2.51 mL AND you can't easily measure out a final solution of 255 mL (and the 68% is just a close 68%), the best you can do is come close to 0.150 M. I realize this is just a practice problem but I wanted to add this information.