Correct option is (c) 55.83
Approximate atomic weight \(= \frac{6.4}{\text{sp.heat}} \)
\(= \frac{6.4}{0.11} \approx 58.2\)
Valency of the element = \(\frac{\text{At. wt.}}{\text{Eq. wt.}} = \frac{58.2}{18.61} = 3\)
Exact atomic weight = Valency x Eq. wt.
= 3 x 18.61
= 55.83