Concentrated nitric acid used in the laboratory work is `68%` nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the denisty of the solution is `1.504" mL"^(-1)` ?

Explanation

Given `68%" HNO"_(3)` by mass that means
68gms mass of `"HNO"_(3)` present in 100 gms of solution.
Molecular weight of `"HNO"_(3)=63`
Number of moles of `"HNO"_(3)=("weight")/("GMW")=(68)/(63)=1.079`
Given density of the solution = 1.504 gm/ml
Volume of solution `= ("Mass of solution")/("density")`
`=(100)/(1.504)=66.5" ml"`
Molarity `=nxx(1000)/("V"(ml))`
`=(1.079xx1000)/(66.5)=16.23" M"`