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