Question

25. 16 g of NaOH is present in 100 ml of an
aqueous solution. Its density is 1.06 g/ml. Mole
fraction of the solute is approximately
A) 25/27
B) 2/27
C) 1/27
D) 26/27
tely​

Answer 1

Answer:

Moles of NaOH =4016 =0.4

Density = 1.06 g/ml

1ml weighs 1.06 gms

100 ml solution weighs 11.06×100=106gms

Therefore, amount of solvent = 106 − 16 = 90 gms

Moles of solvent =1890=5

Now,

Xsol= Mole fraction of solute

               =Moles of solute+moles of solventMoles of solute

Xsol=0.4+50.4=272