Question

Hii

A 1.2% solution (w/v) of NaCl is isotonic with 72% solution (w/v) of glucose. calculate degree of ionisation and van't Hoff factor of NaCl .

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Answer 1

$<b>$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 πN = {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

Answer 2

Answer:

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