Question

If 4 g of \[\ce{NaOH}\] dissolves in 36 g of \[\ce{H2O}\], calculate the mole fraction of each component in the solution. Also, determine the molarity of solution (specific gravity of solution is 1g mL^–1).

Answer 1

Molar mass of \[\ce{NaOH}\] = 40 g/mole

Molar mass of water = 18 g/ mole

Mass of \[\ce{NaOH}\] = 4 g

Number of moles of NaOH = `(4  g)/(40  g)` = 0.1 mol

Number of moles of H₂O = `(36  g)/(18  g)` = 2 mol

Mole fraction of water = `(26  g)/("Number of moles of NaOH" + "Number of moles of"  H_2O)`

Mole fraction of NaOH = `2/(0.1 + 2) = 2/2.1` = 0.95

Number of moles of `2/0.1 = 0.1/2.1` = 0.047

Number of moles of \[\ce{NaOH}\]

Mass of solution = Mass of solute + Mass of solvent

Mass of \[\ce{NaOH}\] + Mass of water = 4 g + 36 g = 40 g

Specific gravity of solution = 1 g/ml

1 litre = 1000 ml volume of solution = 40 ml

∴ 40 ml = 0.04 litre

Molarity = `"Number of moles of solute"/"Volume in litre" = (0.1  "mol")/(0.04  "litre")` = 2.5 M