Question

What is the molality of 10% w/w aquauous solution of sodium hydroxide

Answer 1

10% w/w NaOH means that we have 10 g of NaOH in 100 g of solution.

Number of moles of NaOH present = Mass / Molar Mass = 10/40 = 0.25 mol of NaOH

Mass of water present in a solution = 100-10 = 90 g water =0.090 kg water

Number of moles of water = 90 /18 = 5 mol of water

Molality = Number of moles of solute / kg of solvent = 0.25/0.090 = 2.77 m NaOH

​Total number of moles = 0.25 + 5 = 5.25 moles

Mole fraction of NaOH(xNaOH) = number of moles of NaOH/ Total number of moles = 0.25 / 5.25 = 0.0476

Answer 2

Answer:

10% w/w NaOH means that we have 10 g of NaOH in 100 g of solution.

Number of moles of NaOH present = Mass / Molar Mass = 10/40 = 0.25 mol of NaOH

Mass of water present in a solution = 100-10 = 90 g water =0.090 kg water

Number of moles of water = 90 /18 = 5 mol of water

Molality = Number of moles of solute / kg of solvent = 0.25/0.090 = 2.77 m NaOH

​Total number of moles = 0.25 + 5 = 5.25 moles

Mole fraction of NaOH(xNaOH) = number of moles of NaOH/ Total number of moles = 0.25 / 5.25 = 0.0476

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