If 4 g of NaOH dissovles in 36g of `H_(2)O`, calculate the mole fraction of each component in the solution. (specific gravity of solution is `1g mL^(-1)`).

Explanation

Step-I Calculation of the mole fractions of the components
Mass of NaOH = 40g
Molar mass of NaOH `= 40 g mol^(-1)`
No. of moles of NaOH `= ("Mass")/("Molar mass")=(("4 g"))/(("40 g mol"^(-1)))=0.1` mol
Mass of water = 36 g
Molar mass of water `= 18 "g mol"^(-1)`
No. of moles of water `= ("Mass")/("Molar mass")=(("36 g"))/(("18 g mol"^(-1)))=2.0` mol
Mole fraction of NaOH `= (("0.1 mol"))/((0.1+2.0)"mol")=0.048`
Mole fraction of `H_(2)O=(("2.0 mol"))/((0.1+2.0)"mol")=0.952`.
Step-II Calculation of the molarity of the solution
Mass of the solution = Mass of NaOH + Mass of water
`= (4g + 36 g)=40g`
Volume of the solution `= ("Mass of the solution")/("Specific gravity of the solution")=((40g))/(("1 gm L"^(-1)))=40mL`
Molarity of the solution `(M)=("No. of moles of NaOH")/("Volume of solution in litres")`
`= ((0.1 "mol"))/((40//100L))=2.5 "mol L"^(-1)=2.5M`.