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Question: A 0.6% urea solution would be isotonic with A. \(0.1M\) glucose solution. B. \(0.1M\) \(KCl\)sol...

A 0.6% urea solution would be isotonic with
A. 0.1M0.1M glucose solution.
B. 0.1M0.1M KClKClsolution
C. 0.6%0.6\% glucose solution
D. 0.6%0.6\% NaClNaClsolution.

Explanation

Solution

We know that the meaning of isotonic is that the osmotic pressure of the solutions will be equal. The osmotic pressure of a solution is given by
π=iCRT\Rightarrow \pi = iCRT.
Here, π=\pi = osmotic pressure , i=i = van’t hoff factor , C=C = concentration , R=R = ideal gas constant , T=T = temperature.

Complete step by step solution:
We know that urea is a non electrolyte. The molecular formula of urea is NH2CONH2N{H_2}CON{H_2}. We are given a 0.6%0.6\% urea solution. That means it will contain 0.6g0.6g of urea in 100g100g of the solution. The density of the solution will be 1g/ml1g/ml. The volume of the solution will be 100ml=0.1L100ml = 0.1L The molecular mass of urea is 6060. Concentration of a solution is given by

C=molesvolume(L)=0.6600.1=0.1M \Rightarrow C = \dfrac{{moles}}{{volume(L)}} = \dfrac{{\dfrac{{0.6}}{{60}}}}{{0.1}} = 0.1M.

For non electrolytes like urea the i=i = van’t hoff factor will be one. So the osmotic pressure will be π=iCRT=0.1RT \Rightarrow \pi = iCRT = 0.1RT. Now for a 0.1M0.1M glucose solution, the value of i=i = van’t hoff factor will be one . So the osmotic pressure will be π=iCRT=0.1RT \Rightarrow \pi = iCRT = 0.1RT.
For 0.1M0.1M KClKCl solution the i=i = van’t hoff factor will be two because it is a strong electrolyte. So
π=iCRT=0.2RT\Rightarrow \pi = iCRT = 0.2RT. For 0.6%0.6\% glucose solution, the molecular mass of glucose is 180g180g, the osmotic pressure will be
π=iCRT=0.61800.1RT=0.033RT<0.1RT\Rightarrow \pi = iCRT = \dfrac{{\dfrac{{0.6}}{{180}}}}{{0.1}}RT = 0.033RT < 0.1RT.

For 0.6%0.6\% NaClNaClsolution , the molecular mass of NaClNaCl is 58.5g58.5g. The i=i = van’t hoff factor will be two because it is a strong electrolyte. So

π=iCRT=2×0.658.50.1RT=0.0205RT<0.1RT \Rightarrow \pi = iCRT = 2 \times \dfrac{{\dfrac{{0.6}}{{58.5}}}}{{0.1}}RT = 0.0205RT < 0.1RT.

So from the above explanation and calculation it is clear to us that 0.6%0.6\% urea solution and 0.1M0.1M glucose solution are isotonic with each other because the osmotic pressure is same for them that is π=iCRT=0.1RT \Rightarrow \pi = iCRT = 0.1RT.

So, the correct answer is option A.

Additional information:
Glucose has a molecular formula of C6H12O6{C_6}{H_{12}}{O_6}. It is a type of sugar. Osmotic pressure is a colligative property . The meaning of colligative property is that it will depend upon the number of moles of the solute of the solution.

Note: Always remember that the osmotic pressure of a solution is given by π=iCRT \Rightarrow \pi = iCRT. KClKCl and NaClNaCl are strong electrolytes. They get dissociated completely into their constituent ions. Urea and glucose are not strong electrolytes and the degree of dissociation is very low in urea and glucose.