`4.0 g` of `NaOH` is contained in one decilitre of aqueous solution. Calculate the following in the solution (d of `NaOH` solution `= 1.038 g mL^(-1)`)
a. Mole fraction of `NaOH`
b. Molartiy of `NaOH`
c. Molality of `NaOH`

Explanation

`M = (W_(2) xx 100)/(Mw_(2) xx V_(sol))` (1 decilitre `= 100 mL`)
`= (4 xx 1000)/(40 xx 100) = 1`
`m = (W_(2) xx 1000)/(Mw_(2) xx ("Weight of solution - Weight of solute")`
`W_(1)` = weight of solvent
= Weight of solution - weight of solute
`= V_(sol) xx d_(sol) - W_(2)`
`= 100 xx 1.038 - 4 = 99.8 g`
`:. m = (4 xx 1000)/(40 xx 99.8) = 1.002`
`X_(2) = (n_(2))/(n_(1) + n_(2)) = (W_(2) // Mw_(2))/((W_(1))/(Mw_(1)) + (W_(2))/(Mw_(2)))`
`= (4 // 40)/((99.8)/(18) + (4)/(40)) = (0.1)/(5.54 + 0.1) = 0.0177`