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 = 

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

ii) Molarity =  No of moles of NaOHvolume of solution in l=0.10.1=1 mol/l
Regards 

  • 0
What are you looking for?