Commercial nitric acid comes in a concentration of 16.0 mol/L. density of the soln is 1.42g/ml?

Question

what is the percent W/W of nitric acid in this solution?

how many ml of the concentrated acid have to be taken to prepare a 500g of a solution that is 6.00% (W/W) HNO3?

can you teach me how to compute for this? i really am clueless :((

Answer 1

500 g soln x 6g/100g  x 1 mol/63g x 1L/16 mol

You basically start with 500 g of soln, since for every 100g of soln, there are 6 g of acid, then you look for the number of moles needed, so for every 1 mol of acid, there are 63 grams (1+14+3(16)) of acid, then since commercial acid is is 16 mol/L, then you multiply 1L/16mol. your final answer would be 29.76 mL.

If you check the answer

(29.76mL x 1 L/1000 mL x 16 mol/L x 63 g/mol)/500 mL, you would get approximately 6% (slight difference due to rounding off of final answer).

Btw, you don't need the density to compute it.

Hope I helped you solve it.