1.2 % NaCl solution is isotonic with 7.2 % glucose solution. What will be the van't Hoff factor, i?

Options

A. 0.5
B. 1
C. 2
D. 6

Answer

C

Explanation

For NaCI
`pi_(1)V_(1) = i n_(1) RT = i xx W_(1)/M_(1) RT`
`pi_(1) xx 100/1000 = i xx 1.2 /58.5 xx R xx T `
For Glucose
`pi_(2)V_(2) = 1 n_(2) RT = i xx W_(2)/M_(2) RT`
`pi_(2) xx 0.1 =1 xx 7.2/180 xx R xx T `
(for glucose, i=1 neigher association nor dissociation)
NaCI and Glucose are isotonic (Given)
`:. pi_(1) = pi_(2)`
Dividing equation (i) and (ii)
`(pi_(1) xx 0.1)/(pi_(2) xx 0.1) = i xx (1.2/58.5)/(1 xx 7.2 /180)`
`(i xx 1.2 xx 180)/(7.2 xx 58.5)`
`:. i = (7.2 xx 58.5)/(1.2 xx 180)= 1.95 =2 `