 Comment
Share
Q)

# The specific heat of a metal A is found to be 0.03. 10g of a metal gave on

The specific heat of a metal A is found to be 0.03. 10g of a metal gave on evaporation with nitric acid 18.9g of pure dry nitrate. Calculate the equivalent weight and the exact atomic weight of the metal. Comment
A)
By double decomposition method

$\frac{ Eq mass of the metal}{Eq mass of the mettalic nitrate} = \frac{Mass of the metal}{Mass of metallic nitrate}$

$\therefore \frac{E}{E+Eq mass of the NO_3^-} = \frac{10}{18.9}$

$\therefore \frac{E}{E+(14+16 \times 3} = \frac{10}{18.9}$

$\therefore 18.9 E = 10(E+62)$

$\therefore 18.9 E = 10E + 620$

$\therefore 8.9E = 620$

$\therefore E = 69.66$

Atomic Mass x specific heat = 6.4

Atomic mass = $\frac{6.4}{Specific Heat}$

Atomic mass = $\frac{6.4}{0.03} = 213.33$

Now approx atomic = Equivalent mass X Valency

$\therefore$ Valency = Approx Atomic Mass / Equivalent mass

$\therefore Valency = \frac{213.33}{69.66} = 3.06$

Hence Atomic Mass = Equivalent Mass X Valency

Atomic Mass = $69.66 \times 3 = 208.98 U$