A 1.2% solution (w/v) of NaCl is isotonic with 7.2% solution(w/v) of glucose. Calculate degree of ionization and Van't Hoff factor of NaCl.

Explanation

Let us first calculate oberved molar mass of NaCl.
For isotonic solutions.
` pi (NaCl) = pi ("urea")`
and `" " pi = (w RT)/(Mv)`
Osmotic pressure of NaCl, .
` pi (NaCl) = (1.2RT)/( M xx 0.1)`
Osmotic pressure of urea .
` pi ("urea") = (7.2 RT)/(180 xx 0.1) `
Now , `(1.2RT)/(M xx 0.1) = (7.2 RT)/(180 xx 0.1)`
or `" " M = (180 xx 1.2)/(7.2) = 30`
Normal molar mass of NaCl = 58.5
` i=("Normal molar mass")/("Observed molar mass") = (58.5)/(30) = 19.5`
NaCl dissociates as :
`NaCl hArr Na^(+) + Cl^(-)`
If `alpha` is the degree of dissociation .
`{:(,"Initial moles", 1,0,0),(,"Moles after dissociaiton",1-alpha,alpha,alpha):}`
Total moles after dissociation ` = 1 - alpha + alpha + alpha = 1 + alpha`
`i=("Moles of solute after dissociaiton")/("Normal moles of solute") = (1 + alpha)/(1)`
Now , `" " (1+ alpha)/(1) = 1.95`
` alpha = 1.95 - 1 =0.95`