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

1.66 m, 1.57 M

Explanation

Let Mass of `14% H_(2)SO_(4) ` solution = 100 g
`:.` Mass of `H_(2) SO_(4) = 14 g`
Mass of water = 100 - 14
= 86 g = 0 . 0 86 kg
Volume of Solution `= ("Mass of solution ")/("Density of solution ")`
`= (100 g)/( 1 . 10 g cm^(-3)) = 90 . 91 cm^(3)`
`= 0 .091 L` Number of moles of `H_(2) SO_(4)`
`= (14 g)/( 98 g mol^(-1)) = 0 . 143 mol`
(Molar mass of `H_(2) SO_(4) = 98 g mol^(-1)` )
Molarity `= ("Moles of solute ")/("Volume of solution ")`
`= (0 . 143 mol)/( 0.09 1 L) =(143)/(91) = mol L^(-1)`
= 1.57 M
Molality `= ("Moles of solute ")/("Mass of water in kg ") `
`= (0 . 143 mol)/( 0.086 kg) = (143)/( 86) mol kg^(-1)`
= 1.66 m