Question

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 density of solution is `1.504 g mL^(-1)`?

Answer 1

68% nitric acid by mass means that 68g mass of nitric acid is dissolved in 100 g mass of solution.
Molar of `HNO_(3)=63" g "mol^(-1)`
`therefore68" g of "HNO_(3)j=j(68)/(63)=1*079` mole
Density of solution `=1*504" g "mL^(-1)` (given)
`therefore` Volume of solution
`=("mass")/("Density")=(100)/(1*504)=66*5`mL
`therefore` Molarity of solution
`=("Moles of solute"xx1000)/("Volume of solution in mL")`
`=(1*079xx1000)/(66*5)=16*23`M