Question

Calculate the osmotic pressure of $4.5$ g glucose (Molar mass= $180$) dissolved in $100$ mL of water at $298$K. (Given $R = 0.0821Latmmo{l^{ - 1}}{K^{ - 1}}$)

Answer 1

Hint: Osmotic pressure is a colligative property. Colligative properties are those which depend upon the number of particles of solute in a solution.

Complete step by step answer: Osmosis is the movement of solvent through selectively permeable membrane from a solution of low concentration to high concentration. The minimum pressure that should be applied on the solution side to prevent osmosis is called osmotic pressure. It is a colligative property. The formula for osmotic pressure is,
$\pi = icRT$
Where $\pi $ is the osmotic pressure, i is van't-Hoff factor, c is concentration of solution, R is the gas constant and T is the temperature in Kelvin.
Concentration of solution can be written as,
$c = \dfrac{n}{V} = \dfrac{w}{{MV}}$
Where n is the number of moles of solute, w is the weight of solute in grams , M is the molar mass of solute and V is the volume of solution in litre.
So the equation for osmotic pressure can be rewritten as,
$\pi = \dfrac{{iwRT}}{{MV}}$
Given,
w $ = 4.5$g
$R = 0.0821Latmmo{l^{ - 1}}{K^{ - 1}}$
T $ = 298$ K
M $ = 180g/mol$
V $ = 100mL = 0.1L$
i for glucose is $1$.
We need to find the osmotic pressure. Let’s substitute the values in the above equation.
$\pi = \dfrac{{iwRT}}{{MV}} = \dfrac{{1 \times 4.5 \times 0.0821 \times 298}}{{180 \times 0.1}} = 6.12$ atm.
Hence, osmotic pressure of the solution is $6.12$ atm.

Note:
The value of i depends on the nature of the solute and solvent. For glucose, we don’t need to worry about i. Because glucose does not undergo association or dissociation when dissolved in water. Hence the value of i is equal to unity. But some solutes when dissolved in certain solvents undergo association /dissociation. In that case we need additional factors like degree of association/dissociation and number of particles into which one molecule associate/dissociate to calculate i.