7, A 1.2% solution of NaCl is isotonic with 7.2% solution ofglucose. C

1.	7, A 1.2% solution of NaCl is isotonic with 7.2% solution ofglucose. Calculate the van't Hoff factor of NaCI. [Ans. 1.951
Answer» we have, as percentage of weight/volume = (wt. of solute/volume of solution)× 100 Therefore, wt. of glucose = 7.2 g ; volume of solution = 100 mL For glucose : πexpor πN= {w/(m × V)} × ST [V in litre] As, πexpor πN= (7.2 × 1000 × 0.0821 × T)/(180 × 100) For NaCl πN= {w/(m × V)} × ST = (1.2 × 1000 × 0.0821 × T)/(58.5 × 100) As, two solutions are isotonic and hence, πexpNaCl= πNglucose Therefore, for NaCl : πexp/πN= 1 + α Or, {(7.2 × 1000 × 0.082 × T)/(180 × 100)} × {(58.5 × 100)/(1.2 × 1000 × 0.082 × T)} = 1 + α = i T herefore, α = 0.95 & i = 1.95

7, A 1.2% solution of NaCl is isotonic with 7.2% solution ofglucose. Calculate the van't Hoff factor of NaCI. [Ans. 1.951

Answer»

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

Therefore, wt. of glucose = 7.2 g ; volume of solution = 100 mL

For glucose : πexpor πN= {w/(m × V)} × ST [V in litre]

As, πexpor πN= (7.2 × 1000 × 0.0821 × T)/(180 × 100)

For NaCl πN= {w/(m × V)} × ST

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

As, two solutions are isotonic and hence,

πexpNaCl= πNglucose

Therefore, for NaCl : πexp/πN= 1 + α

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

= i T

herefore, α = 0.95 & i = 1.95