4.0g of NaOH is contained in one decilitre of a solution. Calculate the following in the solution

1)MOLe Fraction of NaOH

2)Molality of NaOH

3)Molarity of NaOH

Density of NaOH solution is 1.038g/cm3

= Mass = density × volume

Mass of 0.1 l NaOH solution = 1.038 g/cm3 ×100 cm3 = 103.8 g

1) 4 g of NaOH is present in 103.8 g of solution.
Therefore solvent is 99.8 g.
No of moles of NaOH = 

440=0.1 mol

No of moles of water = 

99.818=5.44 mol

Total no of moles = 0.1 + 5.544 = 5.644 moles

Mole fraction of NaOH = No of moles of NaOH Total no of moles=0.15.644=0.0177

2) Molality = No of moles of NaOH×1000weight of solvent in g=0.1×100099.8=1.002 mol/kg

3) Molarity = ‚ÄčNo of moles of NaOHvolume of solution in l=0.10.1=1 mol/l

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