1.2% NaCl solution is isotonic with 7.2% glucose solution what will be it's van't Hoff factor??

Both solution are isotonic with each other means the osmotic pressure of both solution would be same

Therefore,

we have, as percentage of weight/volume = (wt. of solute/volume of solution) x 100

So, glucose = 7.2 g ; volume of solution = 100 mL

For glucose: pi exp or pi N = {w/(m x V)} x ST [V in litre]

As, pi exp or pi N = (7.2 x  1000 x  0.0821 x T)/(180 x 100)

For NaCl :  pi N = {w/(m  V)} x ST

= (1.2 x 1000 x 0.0821 x T)/(58.5 x  100)

As, two solutions are isotonic and hence,

PiexpNaCl = PiNglucose

Therefore, for NaCl : Pi exp/Pi N = 1 + alpha

Or, {(7.2 x 1000 x 0.082 x T)/(180 x 100)}  {(58.5 x 100)/(1.2 x  1000 x 0.082 x  T)} = 1 + alpha

= i

Therefore, alpha = 0.95 & i = 1.95