Find the molarity and molality of a `67%` solution of `HNO_(3)` by weight . The density of the solution is 1.50 4 g/ cc and the molecular mass of nitric acid is 63

Answer

32.23 m, 15.99 M

Explanation

Mass of `67% HNO_(3)` solution = 100 g (say)
`:.` Mass of `HNO_(3) = 67 g`
Molar mass of `HNO_(3)`
`= (1 + 14 + 48) g mol ^(-1)`
`= 63 g mol^(-1)`
Moles of `HNO_(3) = ("Mass of "HNO_(3))/("Molar mass of " HNO_(3))`
`= (67 g)/( 63 g mol ^(-1)) = 1.063 ` mol
Mass of Solvent = Mass of solution - Mass of solute
`= 160 g - 67 g`
`= 33 . g = 0.033 kg`
Molality ` = ("Moles of solute" (HNO_(3)))/("Mass of solvent "(kg))`
`= (1.063 mol)/( 0.033 kg) = (1063)/(33) mol kg^(-1)`
= 32.21 m
Volume of solution `=("Mass of solution ")/("Density of solution ")`
`= (100 g)/( 1.504 g cm^(-3)) = 66 . 49 mL`
= 0.0665 L
Molalrity `= ("Moles of solute ")/("Volume of solution (L)")`
`=(1.063 mol)/( 0.0665 L) = (1063)/(66.5) mol L^(-1)`
= 15.9857 M