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)`)

Explanation

`W_B=4.0 g V=(1)/10=0.1L`
Molar mass of `NaOH=40 "g mol"^(-1)`
`n_B=W_B/M_B=(4.0g)/(40"g mol"^(-1))=0.1 `mol
Volume of solution, V = 0.1 L = 100 ml
Mass of solution `="d"xxV " " [D=M/V]`
= (1.038) (100)
= 103.8 g
Mass of solvent = (103.8-4)
= 99.8 g=0.0998 kg.
`:.` Molarity `=("no. of mole of solute")/("Volume of solution in litre")=(0.1 mol)/(0.1L)= 1 "mol L"^(-1)= 1M `
Molality `= ("no. of mole of solute")/("Mass of solventin kg")=(0.1"mol")/(0.0998) = 1.002 "mol kg "^(-1)`