Question

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

Answer 1

Mass of `HNO_(3)` in solution68g
Molar mass of `HNO_(3)` =63g `mol^(-1)`
Mass of solution= 100g
`"Density of solution"=("Mass of solution")/("Density of solution")`
`=((100g))/((1.504g mol^(-1)))=66.5 mL=0.0665 L `
`"Molarity of solution"(M)=("Mass of" HMO_(3)//"Molar mass of" HNO_(3))/("Volume of solution in Litres")`
`=((68g//63gmol^(-1)))/((0.0665L))=16.23 molL^(-1)=16.23 M`