The van't Hoff factor of CaCl2 in a dilute aqueous solution is found to be 2.50. Using i = 1 + (n-1)alpha where n is the number of ions per formula unit, the degree of dissociation alpha is

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The van't Hoff factor of $CaCl X_{2}$ in a dilute aqueous solution is found to be $2.50$ . Using $i = 1 + (n - 1) α$ where $n$ is the number of ions per formula unit, the degree of dissociation $α$ is

CaCl X_{2}

gives

n = 3

ions.

2.50 = 1 + 2 α

, so

α = 1.50/2 = 0.75