What is the relation in the osmotic pressure at 273K when 10g glucose (∏ 1 ) , 10g urea (∏2 ) and 10g of sucrose (∏3) are dissolved in 250ml of water
Answers
Explanation:
Given,
Temperature = 273 K
Mass of glucose = 10g
Mass of urea = 10g
Mass of sucrose = 10g
Volume of solvent (water) = 250 ml
Molar mass of glucose = 180.156 g/mole
Molar mass of urea = 60.06 g/mole
Molar mass of sucrose = 342.296 g/mole
The formula used for osmotic pressure is,
\pi=\frac{w\times R\times T}{M\times V}π=
M×V
w×R×T
where,
\piπ = osmotic pressure
w = given mass
R = gas constant
T = temperature
M = molar mass
V = volume of solvent
From the formula, we conclude that the Osmotic pressure is inversely proportional to the Molar mass, since other values are same for all three compounds.
\pi\propto \frac{1}{M}π∝
M
1
Osmotic pressure of glucose, urea and sucrose is represented as \pi_1π
1
, \pi_2π
2
and \pi_3π
3
respectively.
This means that the lower the molar mass of the compound, the higher will be the osmotic pressure of compound.
So, the relationship between the osmotic pressure of glucose, urea and sucrose is,
\pi_2 > \pi_1 > \pi_3π
2
>π
1
>π
3
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