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

Moles of NaOH = 4g
Number of moles of NaOH `=(4g)/(40 g) = 0.1` mol
Mass of `H_(2)= 36`g
Number of moles of `H_(2)O = (36 g)/(18 g)=2` mol
Mole fraction of NaOH `=(0.1)/(2+0.1) = 0.1/2.1 = 0.047`
Mole fraction of water `=2/(2+0.1)=2/(2.1) = 0.95`
Mass of solution = Mass of water + Mass of NaOH = 36 g + 4g = 40 g
Volume of solution `=40 xx 1 = 40` mL
(Since specific gravity of solution is `1g mL^(-1)`)
Molarity `=("Moles of solute")/("Volume of solution in mL") xx 1000`
`=0.1/40 xx 1000 = 2.5` M