Question

A solution has $0.4\% $ urea and $3.42\% $ sucrose. What will be the osmotic pressure of this solution of ${27^o}C$ ?

Answer 1

Hint: In the given question, we have to calculate the osmotic pressure of the solution when the percentages of components of the solution is given. Osmotic pressure is also a colligative property of the solution. It is related to the phenomenon osmosis. Molecular mass of urea = $60\,gmo{l^{ - 1}}$ and sucrose = $342\,gmo{l^{ - 1}}$ .
Formula used- osmotic pressure $\pi = iCRT$

Complete answer:
Osmosis is the phenomenon which refers to the movement of solvent molecules through a semipermeable membrane. The movement of solvent molecules takes from a region where the concentration of the solute is low to the region where the concentration of solute is high. This results in the establishment of equilibrium between the two sides of the semipermeable membrane and the solute concentration is equal on both sides.
Osmotic pressure is defined as the minimum pressure which is applied to the solution to prevent the flow of the pure solvent across the semipermeable membrane. It is one of the colligative property of the solutions.
Osmotic pressure depends on the concentration of the solute particles in the solution.
The formula for osmotic pressure is $\pi = iCRT$
Where, $\pi $ =osmotic pressure
$i$ = van’t Hoff factor
$C$ = molar concentration of the solute in solution
$R$ = Universal Gas constant = $0.0821\,L\,\,atm\,\,mo{l^{ - 1}}$
$T$ = Temperature in kelvin
According to the question, $0.4\% $ urea means $0.4g$ urea in $100\,ml$
${\text{Number of moles = }}\dfrac{{{\text{Given weight}}}}{{{\text{Molecular mass}}}}$
${\text{Number of moles = }}\dfrac{{{\text{0}}{\text{.4}}}}{{{\text{60}}}}$
Now, the molarity is calculated as,
${\text{Molarity = }}\dfrac{{{\text{Number of moles of solute}}}}{{{\text{Volume of solution (in litres)}}}}$
Substituting the values,
Molarity ${C_1} = \dfrac{{\dfrac{{0.4}}{{60}}}}{{0.1}}$
$ = 0.0666M$
Temperature = ${27^o}C\, = \,300K$
Osmotic pressure = ${\pi _1} = {C_1}RT$
${\pi _1} = 0.666 \times 0.0821\, \times 300$
${\pi _1} = 1.642\,atm$
Similarly, $3.42\% $ of sucrose means $3.42g$ in $100\,ml$
Molarity ${C_2} = \dfrac{{\dfrac{{3.42}}{{342}}}}{{0.1}}$
$ = 0.1M$
Osmotic pressure= ${\pi _2} = {C_2}RT$
${\pi _2} = 0.1 \times 0.0821\, \times 300$
${\pi _2} = 2.463\,atm$
So, the total osmotic pressure $\pi = {\pi _1} + {\pi _2}$
$\pi = 1.642 + 2.463$
$\pi = 4.105atm$
Osmotic pressure of this solution of ${27^{\circ}}C$=$\pi = 4.105atm$

Note:
 It is important to note that the osmotic pressure equation is true for the solution which behaves like an ideal solution. The semipermeable membrane does not allow the movement of solute molecules, it only allows the movement of solvent molecules. Solute concentration is equal on both sides.