Question

A 5.85% (wt/vol) NaCl solution will exert an osmotic pressure closest to which one of the following?
(A) 5.85% (wt/vol) sucrose solution
(B) 5.85% (wt/vol) glucose solution
(C) 2 M sucrose solution
(D) 1 M glucose solution

Answer 1

Hint: Colligative properties of solutions come into picture due to the presence of non-volatile solute in a solution. They only depend on the number of the particles of the solute and not on their nature. Osmotic pressure is also a colligative property.

Complete step by step answer:
In order to solve this question we first need to have an understanding about colligative properties. Let us understand colligative properties and the factors on which they depend.
Colligative properties of solutions come into picture due to the presence of non-volatile solute in a solution. They only depend on the number of the particles of the solute and not on their nature. When we add a non-volatile solute to a solvent, it displaces some solvent molecules at the surface of the solution and hence reduces the vapour pressure of the solvent which is termed as the relative lowering of vapour pressure.
Now, let us understand osmotic pressure. It is defined as the minimum pressure that needs to be applied on a solution so as to prevent the entrance of the solvent into the solution through a semipermeable membrane. It is directly proportional to the concentration of the solution and its temperature, hence
$ \pi =CRT$
Where $ \pi $ is the osmotic pressure, R is the gas constant, T is the temperature in Kelvin and C is the molarity of the solution.
Now, let us solve the question. Since we are dealing with NaCl solution which is a strong electrolyte and hence it completely dissociates, one mole of NaCl will produce two moles of particles in solution. Since colligative properties only depend upon the number of non-volatile solute particles, hence the formula for the osmotic pressure will change:
$ \pi =iCRT$
Where i is called the van’t Hoff factor and it is equal to:
$i=\cfrac { Observed\quad colligative\quad property }{ Calculated\quad colligative\quad property } =\cfrac { Calculated\quad molar\quad mass }{ Observed\quad molar\quad mass } $
For solutes that undergo 100% dissociation, it is equal to the number of ions produced from one molecule of the solute.

Since we get 2 ions namely ${ Cl }^{ - }$ and ${ Na }^{ + }$ from one formula unit of NaCl, therefore i=2.
5.85% (wt/vol) NaCl solution means that 5.85 g of NaCl is present in 100 mL of the solution. Let us
calculate the molarity of such a solution:
$C=\cfrac { 5.8g\times 1000 }{ 58.5g/mol\times 100L } =1\quad M(approximately) $
$\Rightarrow \pi =2RT $
For 2 M sucrose solution the value of osmotic pressure will be (sucrose does not undergo any dissociation or association):
$\pi =2RT $
So, the correct answer is “Option C”.

Note: The van’t Hoff factor is greater than 1 if there is dissociation of the solute in the solution. It is less than 1 if there is association of the solute in the solution. For solutes that undergo 100% dissociation its value is equal to the number of ions produced from one molecule of the solute.