If the apparent degree of ionisation of $K C l$ (Mol. mass = $74.5 g m o l^{- 1}$ ) in water at 290 K is 0.49, the mass of $K C l$ which must be made upto $1 d m^{3}$ (1L) of aqueous solution to have the same osmotic pressure as the 2.7% by mass solution of glucose (density of the solution= 1.2 $g / m l$ ) at that temperature is

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Solution

$π = i C R T$
$= \frac{2.7 \times 1000}{\frac{100}{d} \times 180} R T$
$= \frac{1.49 \times w}{74.5} R T$
$\Rightarrow \frac{2.7 \times 1.2 \times 1000}{100 \times 180}$
$= \frac{1.49 w}{74.5}$
$\Rightarrow w = 9$

For glucose i= 1; for KCl i= 1.49 as $α$ =0.49
So $i = 1 + α = 1.49$

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If the apparent degree of ionisation of KCl (Mol. mass =74.5gmol–1) in water at 290 K is 0.49, the mass of KCl which must be made upto 1dm3 (1L) of aqueous solution to have the same osmotic pressure as the 2.7% by mass solution of glucose (density of the solution= 1.2 g/ml) at that temperature is

π=iCRT =2.7×1000100d×180RT =1.49×w74.5RT ⇒2.7×1.2×1000100×180 =1.49w74.5 ⇒w=9 For glucose i= 1; for KCl i= 1.49 as α =0.49 So i=1+α=1.49

$π = i C R T$
$= \frac{2.7 \times 1000}{\frac{100}{d} \times 180} R T$
$= \frac{1.49 \times w}{74.5} R T$
$\Rightarrow \frac{2.7 \times 1.2 \times 1000}{100 \times 180}$
$= \frac{1.49 w}{74.5}$
$\Rightarrow w = 9$

For glucose i= 1; for KCl i= 1.49 as $α$ =0.49
So $i = 1 + α = 1.49$