1.2% NaCl solution is isotonic with 7.2% glucose solution what will be it's van't Hoff factor??
Answer (1)
17th May, 2019
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
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
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