Please answer very soon
Dear Student
Mass = density volume
Mass of 0.1 l NaOH solution = 1.038 g/cm3 100 cm3 = 103.8 g
i) 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 =
ii) Molarity =
Regards
Mass = density volume
Mass of 0.1 l NaOH solution = 1.038 g/cm3 100 cm3 = 103.8 g
i) 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 =
ii) Molarity =
Regards