Question

A 1.2% solution of NaCl is isotonic with 7.2% solution of glucose. Calculate the van’t Hoff factor of NaCl.

Answer 1

For NaCl solution:

w = 1.2 g, V = 100 ml = 0.1 L

∴ n = `1.2/58.5`    ...(∵ Formula unit mass of NaCl = 58.5)

= 0.0205 moles

∵ nV = π RT

∴ `pi_(NaCl) = (i xx 0.0205 xx R xx T)/0.1`

= 0.205 iRT

For glucose solution:

πV = nRT

or `pi_"glucose" = (7.2/180 xx R xx T)/0.1`    ...(∵ Molecular mass of glucose = 180)

= 0.4 RT

Since the two solutions are isotonic,

`pi_"NaCl" = pi_"glucose"`

or 0.205 i RT = 0.4 RT

or `i = (0.4 RT)/(0.205 RT)`

= 1.95

Hence, the van’t Hoff factor for NaCl is 1.95.