Question

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At 300 Kelvin, 36 gram of glucose present per litre in its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bar at the same temperature, what will be its concentration???

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Answer 1

π₁= 4.98

π₂ = 1.52

C₁ = 36/180

C₂ = ?

Now according to van’t hoff equation

Π = CRT

Putting the values in above equation,we get

4.98 = 36/180 RT ------------------------1

1.52 = C₂RT       ------------------------2

Now dividing equation 2 by 1 ,we get

(C₂ x 180) / 36 = 1.52 / 4.98

or

C₂ = 0.0061

Therefore concentration of 2nd solution is 0.061 M

Answer 2

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Concentration of second soln (C2) = 0.061 M

step-by-step explanation:

Given values,

temperature (T) = 300 K

W(solute) 36 gm

π1 = 4.98 bar

π2 = 1.52 bar

V (soln.) = 1 L

We know that,

π = C × R × T

Now,

Molar mass of glucose ( C6H12O6 )

= 12×6 + 12×1 + 6×16

= 72 + 12 + 96

= 180 gm

=> C1 = 36/180 M

=> 4.98 bar = 36 R T/180 .............. (i)

=> 1.52 bar = C2 R T ................... (ii)

Now,

Dividing eqn. (ii) by eqn. (i)

we get,

C2× 180/36 = 1.52/4.98

=> C2 = 1.52×36/4.98×180

=> C2 = 0.061 M

Thus, the concentration of second solution is 0.061 M.