Question

solution having 20% w/ w of NaOH and density of solution is 1.5 g/ml then find molality and mole fraction of NaOH​

Answer 1

Answer:

molality = 37.87 Molal

mole fraction = 0.683 ≈ 0.7

Explanation:

density of solution = 1.5g/mL

w/w % of NaOH  = 20

mass of NaOH  = (w/w %) / molar mass of NaOH

                         = 20 / 23 + 16 + 1

                         = 20/40 = 0.5 g

mass of NaOH = 0.5 g

Volume = mass / density

             = 0.5 / 1.5

             = 0.33 mL = 3.3 x 10⁻⁴ L

mass of solvent = 3.3 x 10⁻⁴ x1 kg

                          = 3.3 x 10⁻⁴ kg

no of moles = 0.5g /40 = 0.0125 moles

molality = 0.0125 / 3.3 x 10⁻⁴ kg

             = 37.87 Molal

no of moles of solvent = 3.3 x 10⁻¹ g/ 18 = 0.0183 mol

mole fraction = 0.0125 / 0.0183

                      = 0.683 ≈ 0.7