Solution:
SOLUTION
k (CaF2 solution) = 3.86 × 10−5 Ω−1cm−1
k (water solution) = 0.15×10−5 Ω−1cm−1
k (CaF2 alone) = ?
k (CaF2 alone) = k (CaF2 solution) −k (Water solution)
=(3.86 − 0.15) × 10−5Ω−1cm−1
=3.71 × 10−5 Ω−1 cm−1
The specific conductance of CaF2 alone is 3.71×10−5 Ω−1cm−1