Number of moles of NaOH.
`NaOH=(4)/(40)=0.1` mol ` {becausen=("Mass"(g))/("Molar mass"(g "mol"^(-1)))}`
similarly. `H_(2)O=(36)/(18)=2` m ol
Mole fraction of `NaOH, X_(NaOH)=(01)/(0.1+2)=0.0476`
Similarly. `X_(H_(2)O)=(n_(H_(2)O)/(n_(NaOH)+n_(H_(2)O))`
`=(2)/(0.1+2) = 0.9524`
Total mass of solution =mass of solute+mass of solvent
`=4+36=40g`
Volume of solution `=("Mass of solution")/("specific gravity") =(40g)/(1g mL^(-1))=40mL`.
Molariy` =("Moles of solute"xx1000)/("volume of solution"(mL))`
`=(0.1xx1000)/(40)=2.5M`