Question

Aqueous solution of NaOH is marked 10% (w/w). The density of the solution is 1.070g cm⁻³ Calculate i) molarity ii) molality and iii) mole fraction NaOH and water Na=23, H=10, O=16
M=2.675, m=2.77, xNaOH =0.0476, xH₂O =0.9523

Answer 1

Hey Dear,

● Answer -

• Molarity = 2.675 M
• Molality = 2.777 m
• Mole fraction (NaOH) = 0.0476
• Mole fraction (H2O) = 0.9523

● Explanation -

NaOH soln is 10% w/w. That is 10 g NaOH in 90 gm H2O.

(i) Molarity -

Molarity = no of moles of solute / volume of soln

Molarity = (10/40) / (0.1÷1.07)

Molarity = 2.675 M

(ii) Molality -

Molality = no of moles of solute / mass of solvent

Molality = (10/40) / (0.090)

Molality = 2.777 m

(iii) Mole fraction of NaOH -

Mole fraction (NaOH) = moles of NaOH / total moles

Mole fraction (NaOH) = (10/40) / (10/40 + 90/18)

Mole fraction (NaOH) = 0.0476

(iv) Mole fraction of H2O -

Mole fraction (H2O) = moles of H2O / total moles

Mole fraction (H2O) = (90/18) / (10/40 + 90/18)

Mole fraction (H2O) = 0.9523

Hope this helped you.