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Osmotic Pressure - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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  • Osmosis and Osmotic Pressure is considered one of the most asked concept.

  • 20 Questions around this concept.

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200 mL of an aqueous solution of a protein contains its 1.26 g. The osmotic pressure of this solution at 300 K is found to be 2.57\times 10^{-3} bar. The molar mass of protein will be \mathrm{(R = 0.083\ L\ bar\ mol^{-1} K^{-1})}

250 \, \mathrm{ml} aqueous solution of glucose win osmotic pressure 1.5 atm at 25^{\circ} \mathrm{C} is mixed with 420 \, \mathrm{ml} aqueous solution of urea of 3.6 \mathrm{~atm} at 25^{\circ} \mathrm{C}. What will be the osmotic pressure

If 400 \mathrm{~mL} of 0.3 \mathrm{M} urea solution is mixed with 200\, \mathrm{ml} of 0.5 \, \mathrm{M} glucose solutions at 300 \mathrm{~K}.Calculate osmotic pressure

If urea solution at 600 \mathrm{~K} has \mathrm{O . P=3.81 \mathrm{~atm}} and glucose solution at \mathrm{400 \mathrm{~K}} has \mathrm{O . P=2.41 \mathrm{~atm}} . If \mathrm{300 \mathrm{ml}} of first solution and \mathrm{450 \mathrm{~mL}} of second solution are mixed at \mathrm{900 \mathrm{~K}} then calculate O.P of resulting solution at \mathrm{900 \, \mathrm{K}}.
 

What will be the osmotic pressure of 0.8 \mathrm{M} urea aqueous solution at 200 \mathrm{~K}

What will be the molarmass of solute at  {27^{\circ} \mathrm{C}}  solution that contains 3.5 \mathrm{~g}  of solute in  300.0 \mathrm{ml}  of solution with an Osmotic pressure of  \mathrm{400pa}.

Find the strength of the solution if solution is having an osmotic pressure of  4.28 \mathrm{~atm}  at  300 \mathrm{~K}.

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Calculate the osmotic pressure of sucrose solution containing 1.84 grams in 150 ml of solution at 17^{\circ} \mathrm{C}

Concepts Covered - 1

Osmosis and Osmotic Pressure

There are many phenomena which we observe in nature or at home. For example, raw mangoes shrivel when pickled in brine (saltwater); wilted flowers revive when placed in freshwater, blood cells collapse when suspended in saline water, etc. If we look at into these processes we find one thing common in all, that is, all these substances are bound by membranes. These membranes can be of animal or vegetable origin and these occur naturally such as pig’s bladder or parchment or can be synthetic such as cellophane. These membranes appear to be continuous sheets or films, yet they contain a network of submicroscopic holes or pores. Small solvent molecules, like water, can pass through these holes but the passage of bigger molecules like solute is hindered.  Membranes having this kind of properties are known as semipermeable membranes (SPM).


Assume that only solvent molecules can pass through these semipermeable membranes. If this membrane is placed between the solvent and solution as shown in Figure given below, the solvent molecules will flow through the membrane from pure solvent to the solution. This process of flow of the solvent is called osmosis.
The flow will continue till the equilibrium is attained. The flow of the solvent from its side to the solution side across a semipermeable membrane can be stopped if some extra pressure is applied on the solution. This pressure that just stops the flow of solvent is called osmotic pressure of the solution. The flow of solvent from dilute solution to the concentrated solution across a semipermeable membrane is due to osmosis. The important point to be kept in mind is that solvent molecules always flow from lower concentration to higher concentration of solution. The osmotic pressure has been found to depend on the concentration of the solution.

The osmotic pressure of a solution is the excess pressure that must be applied to a solution to prevent osmosis, i.e., to stop the passage of solvent molecules through a semipermeable membrane into the solution. This is illustrated in Figure given above Osmotic pressure is a colligative property as it depends on the number of solute molecules and not on their identity. For dilute solutions, it has been found experimentally that osmotic pressure is proportional to the molarity, C of the solution at a given temperature T. Thus:
\mathrm{\Pi=C R T}
Here Π is the osmotic pressure and R is the gas constant.
\mathrm{\Pi=\left(n_{2} / V\right) R T}
Here V is volume of a solution in litres containing n2 moles of solute. If wgrams of solute, of molar mass, M2 is present in the solution, then n= w2 / Mand we can write,
\mathrm{\Pi V=\frac{\mathbf{W}_{2} R T}{M_{2}}}
\mathrm{Thus, M_{2}=\frac{\mathbf{W}_{2} R T}{\Pi V}}
Thus, knowing the quantities w2, T, Π and V we can calculate the molar mass of the solute.
Measurement of osmotic pressure provides another method of determining molar masses of solutes. This method is widely used to determine molar masses of proteins, polymers and other macromolecules. The osmotic pressure method has the advantage over other methods as pressure measurement is around the room temperature and the molarity of the solution is used instead of molality.
As compared to other colligative properties, its magnitude is large even for very dilute solutions. The technique of osmotic pressure for determination of molar mass of solutes is particularly useful for biomolecules as they are generally not stable at higher temperatures and polymers have poor solubility.

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Osmosis and Osmotic Pressure

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