4 . 0 g of NaOH are contained in one decilitre of a solution. Calculate the molarity and molality of this solution .
(Density of solution ` = 1 . 038 g mL^(-1)`)

Answer

1M, 1.002m

Explanation

`W_(B) = 4 . 0 g , V = 0 . 1 L`
Molar mass of NaOH , `M_(B) = 40 g mol^(-1)`
`n_(B) = (W_(B))/(M_(B)) = (4. 0 g)/( 40 g mol^(-1)) = 0 . 1 mol`
Molarity ` = (n_(B))/(V(L)) = (0 . 1 mol)/( 0 . 1 L)`
` = 1 mol L^(-1) = 1 M`
Vol . of solution v = 0 . 1 L = 100 ML
Density of solution ` d = 1 . 0 38 g m L^(-1)`
`:.` Mass of solution ` = d xx V `
`= (1 . 0 38 g mL^(-1)) xx (100 mL)`
= 103 . 8 g
Mass of solute = 4 . 0 g
`:.` Mass of solvent ` = W_(A)`
= (103 . 8 - 4) g
= 99 . 8 g = 0 . 0 998 log
Molality = `(M_(B))/(W_(A) (kg)) = (0 . 1 mol)/( 0 . 0 998 kg)`