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 =
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 =