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

`rho = M/V`
`V =M/rho`
`=100/(1.038) = 96.33 mL`
Values of solufe = `("Given Mass")/ ("Molar Mass")`
`= 4/40 =0.1`
Molarity `=("Moles of Solute")/(" Volume of. Sol")xx1000`
`= (0.1)/(96.33) xx1000 = 1.0 M`
Mass of solvent =?
`rho = M/V`
Mass of sol `= rho xxV`
`= 1.038 xx100`
`= 103.8`
Mass of solvent = `"sol"^n` - solute
`= 103.8 -4 = 99.89`
Molarity `= ("Moles of Solute")/("Mass of Solvent")xx1000`
`=(0.1)/(99.8) xx1000 =1.002 m`