A solution contains a non-volatile solute of molecular mass \[{{M}_{2}}\]Which of the following can be used to calculate the molecular mass of solute in terms of osmotic pressure?
A. \[{{M}_{2}}=\frac{{{m}_{2}}}{\pi }VRT\]
B. \[{{M}_{2}}=(\frac{{{m}_{2}}}{V})\frac{RT}{\pi }\]
C. \[{{M}_{2}}=(\frac{{{m}_{2}}}{V})\pi RT\]
D, \[{{M}_{2}}=(\frac{{{m}_{2}}}{V})\frac{\pi }{RT}\]

Hint: Osmotic pressure is the force required to stop the osmosis process. Osmotic pressure is directly proportional to the concentration of the solution (morality), solution constant (gas constant), and absolute temperature (temperature in Kelvin).

Complete Step by Step Answer:
When a solute is added to a solvent (generally water) it gives out a solution.
Let us take two solutions in one beaker separated through a semi permeable membrane, one solution containing fewer solute particles (less concentrated or dilute) and the other containing a large number of solute particles (more concentrated).
It is very interesting to know that a semi-permeable membrane only allows the passage of solvent molecules from the solution but not solute. Now the solvent of the solution which is a less concentrated move towards the solution which is highly concentrated through a semi-permeable membrane. And this process is known as osmosis.
As solvent molecules move toward the highly concentrated solution thus we fit the piston on the highly concentrated solution and push it until osmosis stops. Thus, the pressure we apply to stop the osmosis process is equal to the force with which solvent molecules move towards a concentrated solution. This pressure which we applied is known as osmotic pressure.
Osmotic pressure is directly proportional to the force with which osmosis occurs which in turn is directly proportional to the difference between the concentration of one solution and with others. Indirectly osmotic pressure depends on the concentration of solute molecules in one solution as compared to another.
It was first derived by Dutch chemist, Jacobus and given as
\[\pi \text{ }=\text{ }CRT\]
π is osmotic pressure, C is concentration of solute molecules in moles per litre, R is gas constant, and T is temperature in Kelvin.

Where C is equal to the number of moles of solute per litre volume \[\left( n/V \right)\]of solution such as
\[\pi \text{ }=\text{ }\frac{n}{V}\left( RT \right)\]
Where n is the ratio of the given mass of solute to the molecular mass of solute (let’s say m2) which is given in question (M2)
\[\pi \text{ }=\frac{{{m}_{2}}}{{{M}_{2}}}\text{ }\times \text{ }\frac{1}{V}\text{ }\left( RT \right)\]
\[{{M}_{2}}\text{ = }\frac{{{m}_{2}}}{V}\text{ }\times \text{ }\frac{1}{\pi }\text{ }\left( RT \right)\]
\[{{M}_{2}}=(\frac{{{m}_{2}}}{V})\frac{RT}{\pi }\]
Thus, the correct option is B.

Note: It is important to note that equation of osmotic pressure (\[\pi \text{ }=\text{ }CRT\]) is same as gaseous pressure which is \[PV=nRT\], in this equation R is gas constant whose value is same as solution constant R present in osmotic pressure equation (\[0.0821\]).