Question

A solution of sucrose (molar mass = 342g/mol) is prepared by dissolving 68.4g of it per litre of solution, what is osmotic pressure ($R = 0.082Latm{K^{ - 1}}mo{l^{ - 1}}$) at 273K?
(A) 3.92 atm
(B) 4.48 atm
(C) 5.92 atm
(D) 29.4 atm

Answer 1

Hint: Osmotic pressure of a substance is directly related to the number of moles of that substance. Osmosis is independent of the nature of the solute.

Complete step by step answer:
Osmosis is a colligative property. Colligative property is a property that refers to the number of solute particles present in the number of solvent molecules in the solution.
There are four different colligative properties which are relative lowering of vapour pressure, elevation in boiling point, depression in freezing point and osmosis.
Among them, Osmosis is the movement of solvent molecules from the region of low concentration to that of high concentration. The two regions are separated by the semipermeable membrane. This membrane only allows the passage of solvent molecules and not solute molecules. Osmosis is independent of the nature of the chemical species present.
Some solutions have the same osmotic pressure at the same temperature. Such solutions are said to be isotonic solutions. So, when isotonic solutions are kept in contact with each other with semipermeable membranes separating two solutions there will not be any movement of solvent molecules in such cases.
Osmotic pressure is the pressure which is required to stop the osmosis. It is represented by the symbol$\pi $.
\[\pi = CRT\]
Here C = concentration of the solution.
          R = gas constant
          T = Temperature of the solution
The concentration can also be written as $C = \dfrac{n}{V}$
So after substituting the value of C in equation of osmotic pressure then,
$\pi = \dfrac{n}{V}RT$ - - - - - - - (1)

We can use the following formula to find the value of n:
$n = \dfrac{{{\text{weight of the substance}}}}{{{\text{molecular weight of the substance}}}}$

Here Sucrose is used as the solute. 68.4g of sucrose is added to the solvent.
Sucrose has the molecular formula ${{\text{C}}_{{\text{12}}}}{{\text{H}}_{{\text{22}}}}{{\text{O}}_{{\text{11}}}}$. So the molecular weight will be 342g/mol which is mentioned in the question. So accordingly the number of moles will be;
$n = \dfrac{{68.4}}{{342}} = 0.2mol$

The solution prepared has volume, V = 1 litre. The gas constant is mentioned which has the value, $R = 0.082Latm{K^{ - 1}}mo{l^{ - 1}}$. The temperature at which the reaction is taking place, T = 273K.
Thus using the formula\[\pi = CRT\], osmotic pressure can be calculated as follows:
$\pi = \dfrac{{0.2}}{1} \times 0.082 \times 273$
$\pi = 4.4772atm$
$\pi $ is approximately equal to 4.48 atm.
So, the correct answer is “Option B”.

Note: The movement of particles of a solute from a more concentrated to a less concentration medium is considered as diffusion. So diffusion is exactly opposite to that of Osmosis. Colligative properties are only for dilute solutions which are considered to behave as ideal solutions.