Question

​ If relative decrease in vapour pressure is 0.4 for a solution containing 1 mol NaCl in 3 mol of H2O, then % ionization of NaCl is (1) 60% (2) 80% (3) 40% (4) 100%

Answer 1

The relative lowering of vapor pressure = i n2 / (n1+n2)

Where i= van't Hoff factor

n2= moles of solute

n1= moles of solvent.

Put the values and you'll get i= 1. 6.

Now in equation % degree of dissociation = (I-1) /(n-1) × 100. For NaCl, n=2.

Hence, your answer is (1) 60%.