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Standard XII

Chemistry

Osmotic Pressure

If the appare...

Question

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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Similar questions

Q. If the degree of ionization of

K C l

in water at 290 K is 0.86. Calculate the mass of

K C l

which must be made up to

1 d m^{3}

of the aqueous solution to the same osmotic pressure as the 4.0 per solution of glucose at that temperature.

Q. Assertion :

1 M

solution of

K C l

has greater osmotic pressure than

1 M

solution of glucose at the same temperature. Reason: In solution

K C l

dissociates to produce more number of particles.

A solution of glucose (molar mass = 180 g/mol) in water is labelled as 10% (by mass). what would be the molality and molarity of the solution?
(density of solution = 1.2 g/mL)

Q. A

5 %

solution of glucose is isotonic with

1.1 %

solution of KCl at

30^{o} C

. Calculate the degree of ionisation of KCl.

Q. A

0.15 M

aqueous solution of

K C l

exerts an osmotic pressure of

6.8

atm at

310 K

. Calculate the degree of dissociation of

K C l

(R = 0.0821 L i t . a t m K^{- 1} {m o l}^{- 1})

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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 $i = 1 + α = 1.49$