Question

The osmotic pressure in atmospheres of \[10%\] solution of cane sugar at \[69{}^\circ C~\] is?
A. \[724\]
B. \[824\]
C. \[8.21\]
D. \[7.21\]

Answer 1

Hint: Osmotic pressure is the force applied by solvent molecules of one solution to another solution through a semipermeable membrane because of the difference in concentration of solute molecules between two solutions. It was the Dutch scientist, Jacobus who give the relationship between osmotic pressure represented as p and concentration of solute particles in the solution represented as c such as
\[p\text{ }=\text{ }icRT\]
where R is gas constant, i is vol’tHoff factor, and T is the absolute temperature at which process osmosis occurs.

Complete Step by Step Answer:
In the given question, we need to find the osmotic pressure of the \[10%\] cane sugar solution. The \[10%\] cane sugar solution means that the solution (formed with the combination of solvent (liquid generally water) and solute (solid)) containing \[10\text{ }g\] of solute cane sugar in \[100\text{ }g\] of solution and thus, the volume of solution is \[100\text{ }ml\].

Let the osmotic pressure of 10% cane sugar solution is p which is equal to nRT (given in hint).
As cane sugar is non-electrolyte (cannot ionise in solution) thus its van't Hoff factor will be equal to 1. The concentration of solute in a solution is defined as the number of moles of solute per litre solution such as \[\frac{n}{V}\]. But as per the question solute is present in gram \[\left( 10\text{ }g \right)\]and the volume of solution is in millilitre \[\left( 100ml \right)\].

To convert the gram mass of solute in moles divide it with its molecular mass which is equal to \[342\text{ }gmo{{l}^{-1}}\] such as \[\frac{10g}{342\text{ }gmo{{l}^{-1}}}\] gram unit present in the numerator and denominator so both cancelled out and inverse mole present in the denominator so, as it goes to numerator mole inverse change to moles such as:
\[\frac{10}{342}moles\] (n)
To convert \[100\text{ }ml\] solution to litre just divide with 1000 such as
\[\frac{100}{1000}litre\] (V)
To find the concentration number of moles of solute is divided by the volume of solution in litre such as
\[\frac{10}{342}\times \frac{1000}{100}mol/litre\]

Now osmotic pressure of 10% cane sugar solution is given as
\[p\text{ }=\text{ }icRT\]
\[p\text{ }=\text{ }\left( 1 \right)c\text{ }RT\]
\[p=\frac{10}{342}\times \frac{1000}{100}(RT)\]; where R isa gas constant whose value is \[0.082\]and the temperature of the solution is given in Celcius
\[(69{}^\circ C)\]but we need absolute temperature so, first convert it to kelvin by adding 273 such as
\[69\text{ }+\text{ }273\text{ }=\text{ }342\text{ }K\]
So,
\[p=\frac{10}{342}\times \frac{1000}{100}(0.082\times 342)\]
Solving it we get
\[p\text{ }=\text{ }8.21\text{ }atm\]
Thus, the correct option is C.

Note: The movement of solvent molecules from low concentrated solution (less number of solute molecules) to a highly concentrated solution (a large number of solute particles) with some force through a semipermeable membrane and this process is known as osmosis. If we applied pressure against the force of the solvent molecule (less concentrated solution) to stop the osmosis process that pressure is osmotic pressure. The osmotic pressure of the solvent molecule of high concentrated solution will be very negligible as compared to the osmotic pressure of the less concentrated solution. In this question, the osmotic pressure of cane sugar solution is \[8.21atm.\]. So applying this amount of pressure will stop the osmosis process and if we applied more than this pressure then reverse osmosis will take place.