Question

The observed osmotic pressure for a 0.01 M solution of $Fe{{(N{{H}_{4}})}_{2}}{{(S{{O}_{4}})}_{2}}$ to ${{25}^{\circ }}C$ is 10.8 atm. The expected and experimental (observed) values of van’t Hoff factor (i) will be respectively.
[ R = 0.082 L atm / K mol ]
(A) 5 and 4.42
(B) 4 and 4.00
(C) 5 and 3.42
(D) 3 and 5.42

Answer 1

Hint: Write the balanced equation for the dissociation of $Fe{{(N{{H}_{4}})}_{2}}{{(S{{O}_{4}})}_{2}}$ and count the total charge on all the compounds, this will give the expected value of (i). And for the experimental value use the formula $i=\dfrac{P}{CRT}$, where P is the pressure, C is the concentration, R is the gas constant, and T is the temperature.

Complete step by step solution:
For some compounds, the expected and experimental value of (i) van’t Hoff factor is not the same.
So for $Fe{{(N{{H}_{4}})}_{2}}{{(S{{O}_{4}})}_{2}}$ the expected value of (i) will be:
We have to write the equation in which this compound splits into ions, the reaction will be:
$Fe{{(N{{H}_{4}})}_{2}}{{(S{{O}_{4}})}_{2}}\to 2NH_{4}^{+}+F{{e}^{2+}}+2SO_{4}^{2-}$
So, 1 mole of the reactant gives 5 moles of the product.
So the expected value of (i) is 5.
Now for the experimental value of (i),
We can apply the formula of osmotic pressure for calculating the value of (i),
The formula is:
 $P=iCRT$
This formula can be written as:
$i=\dfrac{P}{CRT}$
Where P is the pressure, C is the concentration, R is the gas constant, and T is the temperature.
So, in the question P is 10.8 atm, C is 0.01 M, R is 0.082 L atm / K mol, and T is ${{25}^{\circ }}C$ which is equal to 298 K.
Now, putting all these in the formula, we get:
$i=\dfrac{10.8}{0.01\text{ x 0}\text{.082 x 298}}=4.42$
So, the experimental (i) is 4.42.

Therefore, the correct answer is an option (A)- 5 and 4.42.

Note: The other formula that can be used to calculate the van’t Hoff factor (i) is by dividing the normal molecular mass by the abnormal molecular mass of the compound. The temperature must be taken in Kelvin.