Calculate molarity and molality of solution prepared by mixing equal volumes of ` 30%` by mass of `H_(2) SO_(4)` (density = 1 .218 g `mL^(-1)`) and`70%` by mass of `H_(2) SO_(4)` (density `= 1 . 610 g mL^(-1)` )

Answer

`7.617 M, 11.40 m`

Explanation

Let 100 mL of `30% H_(2) SO_(4) ` is mixed with 100
mL of ` 70% H_(2) SO_(4)`
Mass of 100 mL of `30% H_(2) SO_(4)`
`= 100 xx 1 . 218 = 121 . 8 g `
Mass of `H_(2) SO_(4) ` in this solution,
`W_(1) = (121 . 8 xx 30)/( 100) = 36 . 54 g`
Mass of 100 mL of ` 70% H_(2) SO_(4)`
`= 100 xx 1. 610 = 161 . 0 g`
Mass of `H_(2) SO_(4) ` in this solution
`W_(2) = (161.0 xx 70)/(100) = 112 . 7 g`
Total mass of `H_(2) SO_(4)`
`W_(B) = w_(1) + w_(2)`
`= (36 . 54 + 112 . 7) g`
= 149 . 24 g
Molar mass of `H_(2) SO_(4) , M_(B) = 98 g mol^(-1)`
Number of moles of
`H_(2) SO_(4) , n_(B) = (149 . 24 g)/( 98 g mol^(-1))`
=1 . 523 mol
Total Volume of solution = 100 mL + 100 mL
= 200 mL = 0 . 2 L
Molarity = `(n_(B))/("Volume of Solution (L)") = (1 . 523 mol)/( 0 . 2 L)`
`= 7 . 614 mol L^(-1)`
= 7 . 614 M
Total Mass of solution = 121 . 8 g + 161 g
= 282 . 8 g
Total Mass of Solvent `(H_(2) SO_(4))`
=149 . 24 g
`:.` Mass of Solvent (Water)
`= (282 . 8 - 149 . 24) g `
`= 133 . 56 g = 0 . 1336 kg`
Molality ` = (n_(B))/("Mass of solvent " (kg))`
`= (1 . 523 mol)/( 0 . 1336 kg) = 11 . 40 mol kg ^(-1)`
= 11 . 40 m