Question

# Van't Hoff factor of $$Hg_2Cl_2$$ in its aqueous solution will be ($$Hg_2Cl_2$$ is $$80\%$$ ionized in the solution):

A
1.6
B
2.6
C
3.6
D
4.6

Solution

## The correct option is B $$2.6$$$$Hg_2Cl_2\rightleftharpoons Hg^{2+}_2 + 2Cl^-$$Here, $$n=3$$Now,$$\alpha = \dfrac{i-1}{n-1}$$$$0.8=\dfrac{i-1}{3-1}$$$$i=2.6$$Chemistry

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