Question

If NaCl is added to the water and the degree of dissociation is 90%. What is the Van’t Hoff factor?

• A. 1.5
• B. 1.8
• C. 1.9
• D. 0.9

Answer 1

The correct option is C 1.9

Degree of dissociation is fraction of moles dissociated per 1 mole of substance taken. If 1 mole of NaCl is taken 90% of it dissociates, mems 0.9 moles NaCl dissociates.
$$\begin{array}{cccc}Nacl& \to & N{a}^{+}+& c{l}^{-}\\ 1& & 0& 0\\ 1-0.9& & 0.9& 0.9\\ \left(0.1\right)& & & \end{array}$$
Total no. of moles after dissociation =
$$\begin{array}{ccc}0.1+& 0.9+& 0.9\\ \downarrow & \downarrow & \downarrow \\ \left(Nacl\right)& \left(N{a}^{+}\right)& \left(C{l}^{-}\right)\end{array}$$
= 1.9 moles
So Van’t Hoff factor =$\frac{finalno.of.Particles}{Intialno.ofparticles}=\frac{1.9}{1}$
(i) = 1.9.