Question

A 0.1 molar solution of NaCl is found to be isotonic with 1% urea solution. Calculate
(a) Van't Hoff factor
(b) Degree of dissociation of sodium chloride.
Assume the density of 1% urea equal to $\text{1 g c}{{\text{m}}^{-3}}$.
A. (a) 1.667, (b) 0.667
B. (a) 1.82, (b) 0.69
C. (a) 2.01, (b) 0.78
D. (a) 2.12, (b) 0.94

Answer 1

Hint: For this problem, we have to use the formula of the osmotic pressure of the solute to calculate the degree of dissociation that is $\dfrac{\text{n}}{\text{V}}\pi \text{RT}$. Here, n is the number of moles, V is the volume of the solution, R is known as the gas constant and T is the temperature.

Complete step by step solution:
- In the given question, we have to find the value of Van't Hoff factor and degree of dissociation of the sodium chloride.
- Now, it is given in the question, that the sodium chloride and urea are isotonic which means that the concentration of both the solutions is equimolar.
- Firstly, we have to find the concentration of sodium chloride i.e.
$\text{NaCl }\rightleftharpoons \text{ N}{{\text{a}}^{+}}\text{ + C}{{\text{l}}^{-}}$
- Initially the concentration of sodium chloride will be $\text{C(1-}\alpha \text{)}$ and concentration of sodium ion and chloride ion will be $\text{C}\alpha $.
- So, the total concentration will be $\text{C(1 + }\alpha )$ or $\text{0}\text{.1(1 + }\alpha )$.
- So, we will apply the formula of the osmotic pressure of both solution by keeping them equal i.e.
$\dfrac{{{\text{n}}_{\text{NaCl}}}}{{{\text{V}}_{\text{NaCl}}}}\pi \text{RT = }\dfrac{{{\text{n}}_{\text{Urea}}}}{{{\text{V}}_{\text{Urea}}}}\pi \text{RT}$
- Here, we will know the value of n and v only because the value of R and T is the same for both and can cancel out each other.
$\dfrac{0.01(1\text{ + }\alpha \text{)}}{1}\pi \text{RT = }\dfrac{1/60}{0.1}\pi \text{RT}$
$\dfrac{0.01(1\text{ + }\alpha \text{)}}{1}\pi \text{RT = }\dfrac{1/60}{0.1}\pi \text{RT}$
$\alpha \text{ = }\dfrac{1}{60\text{ }\times \text{ 0}\text{.1 }\times \text{ 0}\text{.1}}\ -1$
$\alpha \text{ = }1.667\ -1\text{ = 0}\text{.667}$
- Now, to calculate the Van't Hoff factor we will use the formula:
$\implies \dfrac{\text{C(1-}\alpha \text{)}}{\text{C}}$
$\implies \dfrac{\text{0}\text{.1(1 - 0}\text{.667)}}{0.1}\text{ = 1}\text{.667}$

Therefore, option A is the correct answer.

Note: The Van't Hoff factor is also defined as the ratio of the actual concentration of the substance to the concentration of the substance which is calculated by using the mass. That's why it depends on the concentration of the solute.