Question

Molar conductivity for NaCl,HCl and NaAc are 126.4,425.9 and 91.0 Scm2/mol reapectively calculate molar conductivity for HAc

Answer 1

you can calculate from this above calculation this is a very basic question

Answer 2

Answer : The molar conductivity for HAc is, $390.5Scm^2/mol$

Solution :

Using Kohlorausch's law,

$\Lambda^o_{HAc}=\lambda^o_{H^+}+\lambda^o_{Ac^-}$   ......(1)

$\Lambda^o_{NaCl}=\lambda^o_{Na^+}+\lambda^o_{Cl^-}$   ......(2)

$\Lambda^o_{HCl}=\lambda^o_{H^+}+\lambda^o_{Cl^-}$     ........(3)

$\Lambda^o_{NaAc}=\lambda^o_{Na^+}+\lambda^o_{Ac^-}$    .........(4)

Now adding equation 3 & 4 and then subtracting equation 2, we get the equation 1.

$\Lambda^o_{HAc}=\Lambda^o_{HCl}+\Lambda^o_{NaAc}-\Lambda^o_{NaCl}$

$\Lambda^o_{HAc}=(425.9+91.0)-126.4=390.5Scm^2/mol$

Therefore, the molar conductivity for HAc is, $390.5Scm^2/mol$