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

Explanation

Mass of NaOH = 4g Number of moles of NaOH `= (4g)/(40 g) = 0.1 mol`
Mass of `H_(2)O = 36 g` Number of moles of `= (36 g)/(18 g) = 2 mol`
Mole fraction of water `= ("Numer of moles of " H_(2)O)/("No. of moles of water + No. of moles of NaOH") = (2)/(2 + 0.1) = (2)/(2.1) = 0.95`
Mole fraction of NaOH `= ("Numer of moles of NaOH")/("No. of moles of NaOH + No. of moles of water") = (0.1)/(2 + 0.1) = (0.1)/(2.1) = 0.047`
Mass of solution = mass of water + mass of NaOH = 36 g + 4 g = 40 g
VOlume of solution `= 40 xx 1 = 40 mL`
(Since specific gravity of solution is `= 1 g mL^(-1)`)
Molarity of solution `= ("Number of moles of soluble")/("Volume of solution in litre") = ("0.1 mol NaOH)/(0.04L) = 2.5 M`