Question

The osmotic pressure of a $5\%$ cane sugar at $150^\circ C$ is:
A) $4atm$
B) $3.4atm$
C) ${\text{3}}{\text{.55}}atm$
D) ${\text{5}}atm$

Answer 1

We must know that the osmosis is nothing but solvent molecules pass through semipermeable membranes from the solution of lower concentration to a solution of higher concentration. This semipermeable plays a vital role in passing solvents. The concentration of one side decreases and pressure differences also arise in between the compartments. This kind of pressure difference pushes back pure solvent to the solution side through the semipermeable membrane until equilibrium is established. If equilibrium is attained, the rate of flow of solvents in both lower and higher concentration solutions is equal. At the equilibrium, the pressure difference of the solution is known as osmotic pressure.

Formula used: We must know that the osmotic pressure is directly proportional to the product of the temperature of the solution and molar concentration of the solute.
$\pi \propto {\text{CT}}$
$\pi = RC{\text{T}}$
We replace the above equation by using $C = \dfrac{n}{V}$ and $n = \dfrac{w}{m}$
Hence, $\pi = \;\dfrac{w}{{Vm}}R{\text{T}}$
Here $\pi$ is the osmotic pressure of the solution.
$C$ is the molar concentration of the solute.
${\text{T}}$ is the temperature of the solution in kelvin.
$R$ is the gas constant.
$V$ is the volume of the solution in litre.
$n$ is the number of moles of the solute.
$w$ is the given weight of the solution.
$m$ is the molecular mass of the solute.
Temperature in kelvin = $273$ + temperature in Celsius.

Complete step by step answer:
We must remember that the osmotic pressure is defined as in semipermeable membrane pressure must be applied to the solution to stop influx of the solvent.
Given,
Temperature of the solution is $150^\circ C$ .
We convert temperature in kelvin = $273$ + temperature in Celsius.
Now we can substitute the given value we get,
$= 273{\text{ + 150}}$
On simplification we get,
Temperature in kelvin $= 423K$
Gas constant $R$ = $8.21 \times {\text{1}}{{\text{0}}^{{\text{ - 2}}}}$
The molecular mass of sucrose is $342g$.
The weight of the sucrose in given solution is $5g$
We consider, the volume of the solution is $100ml$ or $0.1L$
Because the weight of the solute is given in percentage of the solution.
Osmotic pressure, $\pi = RC{\text{T}}$
For the given information, we replace by osmotic pressure equation as,
$\pi = \dfrac{w}{{Vm}}R{\text{T}}$
Now we can substitute the known values we get,
$\pi = \dfrac{5}{{342}} \times \dfrac{1}{{0.1}} \times 8.21 \times {10^{ - 2}} \times 423$
$\Rightarrow \pi = \dfrac{5}{{34.2}} \times 34.7283$
On simplification we get,
$\pi = {\text{5atm}}$
Therefore, osmotic pressure is $5{\text{atm}}$.

So, the correct answer is Option D.

Note: We need to remember that the osmotic pressure is represented by the symbol $\pi$ . Osmotic pressure unit is $atm$ . If two solutions having the same osmotic pressure at the given temperature is called isotonic solutions. The net solvent flow in both directions of this isotonic solution is zero. If the concentration of solution is dilute means, that solution is called a hypotonic solution. The concentration of solution increases means that solution is called a hypertonic solution.