# Calculating molar conductance of CaCl2

Ok, so the following question is given in my text book.

Of the molar conductance value of $\ce{Ca^2+}$ and $\ce{Cl^-}$ at infinite dilution are respectively $118.88\times10^{-4}$ and $77.33\times10^{-4}$ then that of $\ce{CaCl2}$ is (all have same unit)

So what I did was I added the molar conductivity of $\ce{Ca^2+}$ and added $2\times$ molar conductivity of $\ce{Cl^-}$. I got the answer. It was one of the options. So I basically wanted to know: Did I do it correctly and is my process correct? And if I did it wrong then what is the correct process?

Yes, the process is correct, but do you understand why?

• As per my knowledge there is one mole of $Ca$ and 2 mole of $Cl$ in $CaCl_2$ that's why am i correct Apr 12, 2014 at 3:54
– LDC3
Apr 12, 2014 at 3:58

You are calculating limiting molar conductivity

limiting molar conductivity = V+ A+ + V- A-

V+ = number of cations per formula unit of the electrolyte

V- = number of anions per formula unit of the electrolyte

A+ = molar conductivity of the cations

A- = molar conductivity of the anions

So , your process is correct.