Question
Thu September 12, 2013 By: Mudita Agrawal
 

Given 4% NaOH solution by mass with density 1.2g per ml find: 1.Molarity 2.Molality 3.Mole fraction of NaOH

Expert Reply
Sun September 15, 2013
Molality means no. of moles of solute per 1000 g or per kg of solvent.Molar mass of NaOH = 23+1+16 = 40 g/mol. So, 4% NaOH aq. means 40 g or 1 mole NaOH in 1000 ml (=1000 g or 1kg) water. Molatity = 1 mol solute / 1kg solvent =1 mol/kg.Volume of solution = mass of solution (40+1000 = 1040 g) /density of 1.2 g per cc = 866.7 cc, which has 1 mole solute in it. Therefore no. of moles of solute per 1000 cc or 1 litre solution = 1000 x1 / 866.7 = 1.153 mole per litre.Mole fraction of NaOH = No. of moles of NaOH / Total moles = 1.0 / (1.0 + 1000 g /18 g per mole) = 1.0 /(1 + 55.55) = 0.017 (it is unitless). Mole fraction of water = 55.55 /56.55 = 0.982. Also, the sum of mole fractions is always 1.00. Rememer, molality is a mass/ /maa unit. So, it the best unit because it is temperature independent whereas molarity is mass /volume unit which is dependent on temp.
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