# 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 =

No of moles of water =

Total no of moles = 0.1 + 5.544 = 5.644 moles

Mole fraction of NaOH =

2) Molality =

3) Molarity = ​

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