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A 1.2% solution of NaCl is isotonic with 7.2% solution of glucose. Calculate the value of Vant Hoff factor of NaCl.

A 1.2% solution of NaCl is isotonic with 7.2% solution of glucose. Calculate the value of Vant Hoff factor of NaCl.

Grade:12

2 Answers

Arun
25750 Points
6 years ago

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 : πexp or π= {w/(m × V)} × ST [V in litre]

As, πexp or 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

Regards

Arun (askIITians forum expert)

Ajeet Tiwari
askIITians Faculty 86 Points
3 years ago
hello students
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

therefore, α = 0.95 & i = 1.95

Thankyou and Regads

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