Question

80 grams of NaOH is dissolved in 500 gram of water.calculate 1)molality of the solution and 2)mole fraction of NaOH in the solution.
(atomic mass H=1,O=16,Na=23)

Answer 1

molality = (weight/gram molecular weight)×(1/kg)
=(80 g/40 g/mol)×(1/.500 kg)
= 1.0 mol/kg
hence the molality of the solution is 1.0 mol/kg

Answer 2

Molarity is the concentration by volume of the solution. = moles/litre
Molality is the concentration by weight of the solvent = moles/kg.  This is an easier and convenient measure, as it is easier to measure the weight than volume.

moles of NaOH in 80 grams = 80/(24+1+16) = 2

Molality = moles of NaOH / weight of  the solvent in kg
             = 2 / [ 500/1000  ]   = 4 moles/kg
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molar mass of water = 18
mole fraction of NaOH :
  number of moles of NaOH in the solution = 80 grams/40 grams = 2
  number of moles of water in 500 grams = 500/18 = 250/9

molar fraction =  2 /[ 250/9 + 2] = 9 /134