Question

If 1 mole of NaCl and 1 mole $AlC{l}_3$ is added to the water. If degree of dissociation of NaCl is 90% and degree of dissociation of $AlC{l}_3$ is 80%. What is the Van’t Hoff factor is?

• A. 1.3
• B. 3.95
• C. 2.65
• D. 4.2

Answer 1

The correct option is C 2.65

For calculating Van’t Hoff factor 1 we have to know the final number of. particles in the solution.
As the degree of dissociation of salts are given we can easily find the no. of particles.
As same way we did in the last question
$$\begin{array}{cccc}NaCl& \to & N{a}^{+}+& C{l}^{-}\\ 1& & 0& 0\\ 1-0.9& & 0.9& 0.9\\ \left(0.1\right)& & & \end{array}$$
$$\begin{array}{cccc}AlC{l}_3& \to & Al+& 3c{l}^{-}\\ 1& & 0& 0\\ 1-0.8& & 0.8& 3\times 0.8\\ \left(0.2\right)& & & 2.4\end{array}$$
We have takne 1 mole of NaCl and $AlC{l}_3$ each.
0.9 moles of NaCl dissociated giving 0.9 moles of $N{a}^{+}$ and $C{l}^{-}$ each and learning 0.1 mole of undissociated NaCl is the solution. 0.8 moles of $AlC{l}_3$ dissociated gives 0.8 moles of $A{l}_3$ + and 2.4 moles of $C{l}^{-}$ ions and learning 0.2 moles of $AlC{l}_3$ undissociated.
So intial no. of particles taken =
$$\begin{array}{ccc}1+& 1& =2\\ \downarrow & \downarrow & \\ \left(NaCl\right)& \left(Alc{l}_3\right)& \end{array}$$
Final no. of all particles = (0.1 + 0.9 + 0.9) + (0.2 + 0.8 + 2.4)
= 1.9 + 3.4 = 5.3
Van’t Hoff factor $\frac{5.3}{2}=2.65$