`4 g` of `NaOH` are present in `0.1 dm^(3)` solution have
(a) mole fraction of `NaOH`,
(b) molality of `NaOH` solution,
(c ) molarity of `NaOH` solution,
(d) normality of `NaOH` solution.

Explanation

Volume of solution `= 0.1 dm^(3)`
`=(1)/(10)"litre"= 100 mL`
Weight of solution `= 100xx1.038= 103.8 g`
Weight of solution `=100xx1.038= 103.8 g`
Weight of water`=` weight of solution `-` weight of `NaOH`
`=103.8-4= 99.8 g`
(a) Mole fraction of `NaOH={((4)/(90))/((4)/(40)+(99.8)/(18))}= 0.018`
(b)Molality of solution `=("Moles of "NaOH)/("Weight of water")xx1000`
`=(4xx1000)/(40xx99.8)= 1.002 m`
(c ) Molarity of `NaOH` solution`=("Moles of" NaOH)/("Volume in litre")`
`=(4xx1000)/(40xx100)= 1M`
(d) Normality of `NaOH` solution
`=("Equivalent of "NaOH)/("Volume in litre")=(4xx1000)/(40xx100)= 1N`