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

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» <p>we have, as percentage of weight/volume = (wt. of solute/volume of solution)× 100</p><p>Therefore, wt. of glucose = 7.2 g ; volume of solution = 100 mL</p><p>For glucose : πexpor πN= {w/(m × V)} × ST [V in litre]</p><p>As, πexpor πN= (7.2 × 1000 × 0.0821 × T)/(180 × 100)</p><p>For NaCl πN= {w/(m × V)} × ST</p><p>= (1.2 × 1000 × 0.0821 × T)/(58.5 × 100)</p><p>As, two solutions are isotonic and hence,</p><p>πexpNaCl= πNglucose</p><p>Therefore, for NaCl : πexp/πN= 1 + α</p><p>Or, {(7.2 × 1000 × 0.082 × T)/(180 × 100)} × {(58.5 × 100)/(1.2 × 1000 × 0.082 × T)} = 1 + α</p><p>= i T</p><p>herefore, α = 0.95 & i = 1.95</p>