1. 1.2 solution(W/V) of Nacl is isotonic with 7.2% solution(W/V) of glucose (molar mass 180 g mol-1. the degree of dissociation of nacl solution is____
2. 'x'g of non electrolyte compound (molar mass 200) are dissolved in 1.0 Lof 0.05 M Nacl solution. the osmotic pressure of thhis solution will be 4.92 atm at 27°C.Assume complete dissociation of Nacl and ideal behaviour of this solution(R=0.082 L mol-1 k-1). the value of X is ___
111. φ( osmaotic pressure) of NaCl= osmotic pressure of glucose
1.2*R*T/M *0.1 = 7.2*R*T/ 180*0.1
M=30.
this is the observed mass.
i = normal mass/ observ. mass
i=58.5/ 30
i=1.95
degree of dissociation of NaCl (α)=1.95-1/2-1
α = 0.95
% of dissociation= α*100
% of dissociation= 0.95*100
% of dissociation =95%
1for question 2, take d equation pi=icRT
here i is 2 because of complete dissociation(2 ions Na+ and Cl- are formed).
c is x/200*100/1
put the values in the formula to get x.
i think u'll be able to solve d same now.
