Question

$1.0 g$ of a metallic carbonate left on ignition $0.56 g$ of its oxide. What is the atomic weight of the metal if its specific heat is $0 \cdot 155 cal / g ^{\circ} C$?

Answer 1

SOLUTION : Atomic weight (approx.) $=\frac{6 \cdot 4}{\text { sp. heat }}=\frac{6 \cdot 4}{0 \cdot 155}=41 \cdot 29$

Calculation of equivalent weight

$\begin{array}{l}\frac{\text { wt. of metallic carbonate }}{\text { wt. of metallic oxide }}=\frac{\text { eq. wt. of metallic carbonate }}{\text { eq. wt. of metallic oxide }} \\\qquad \frac{1 \cdot 0}{0.56}=\frac{E+30}{E+8} \quad \text { (eq. wt. of } CO _{3}^{2-}=\frac{60}{2}=30 . \text { ) }\end{array}$

or, $\quad \frac{1 \cdot 0-0 \cdot 56}{0 \cdot 56}=\frac{E+30-E-8}{E+8}$

or, $\quad \frac{44}{56}=\frac{22}{E+8}, \quad \therefore \quad E+8=28, \quad \therefore \quad E=20 \cdot 0$

Valency $=\frac{\text { rough at. wt. }}{\text { exact eq. wt. }}=\frac{41 \cdot 29}{20} \approx 2, \therefore$ Exact atomic weight $=20 \cdot 0 \times 2=40 \cdot 0$.