Question

If \[3\text{ }g\] of glucose (molecular mass 180) is dissolved in \[60\text{ }g\] of water at \[15{}^\circ C\], then the osmotic pressure of this solution will be :
A. \[0.34\text{ }atm\]
B. \[0.65\text{ }atm\]
C. \[6.57\text{ }atm\]
D. \[5.57\text{ }atm\]

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’t Hoff 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 glucose solution (formed with the combination of solvent (liquid generally water) and solute (solid)) containing \[3\text{ }g\]of glucose solute and 60 g water solvent.
Let the osmotic pressure of glucose solution is p which is equal to icRT (given in hint).
As glucose is non-electrolyte (cannot ionize in solution) thus its vol’t Hoff factor (i) will be equal to 1. The concentration of solute in a solution is defined as the number of moles of solute per litre volume solution such as \[n/V\]. But as per the question solute is present in gram (3g) and the weight of solution is in \[60\text{ }g\].
To convert the gram mass of solute in moles (n) divide its given weight (\[3\text{ }g\])with its molecular mass which is given, 180 such as
\[\frac{3}{180}\] (n)
Also we need volume in litre so we need to divide given mass (given in gram, \[60\text{ }g\]) of solvent (water) with 1000 such as
\[\frac{60}{1000}\]
To find the concentration, number of moles of solute (n) is divided by the volume of solution in litre (V) such as
\[\frac{3}{180}\times \frac{1000}{60}\]
Now, osmotic pressure of glucose solution is given as
\[p\text{ }=\text{ }icRT\]
\[p\text{ }=\text{ }\left( 1 \right)c\text{ }RT\]
\[p=\frac{3}{180}\times \frac{1000}{60}(RT)\]
Where R is a gas constant whose value is \[0.082\]and the temperature of the solution is given in Celcius
\[\left( 15\text{ }{}^\circ C \right)\]but we need absolute temperature so, first convert it to kelvin by adding 273 such as
\[15\text{ }+\text{ }273\text{ }=\text{ }288\text{ }K\]
So,
\[p=\frac{3}{180}\times \frac{1000}{60}(0.082\times 288)\]
\[p\text{ }=\text{ }0.28\text{ }\times \text{ }\left( 0.0821\text{ }\times \text{ }288 \right)\]
Solving it we get
\[p\text{ }=\text{ }6.57\text{ }atm\]
Thus, the correct option is C.

Note: The movement of solvent molecules from a 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 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 the glucose solution is \[6.57\text{ }atm\] so, applying this amount of pressure will stop the osmosis process and if we apply more than this pressure then reverse osmosis will take place.