1.878g of a metal bromide when heated in a stream of hydrogen chloride gas is completely converted to the chloride weighing 1.00g.

Question

$1.878 g$ of a metal bromide when heated in a stream of hydrogen chloride gas is completely converted to the chloride weighing $1 \cdot 00 g$. The sp. heat of the metal is $0 \cdot 14 cal / g ^{\circ} C$. Calculate the molecular weight of the metal bromide $( Br =80, Cl =35 \cdot 5)$.

Answer 1

SOLUTION : In order to find out the molecular weight of metal bromide it is essential to evaluate the atomic weight and valency of the metal.

By Dulong and Petit's law

$\text { at. wt. (approx.) }=\frac{6 \cdot 4}{\text { sp. heat }}=\frac{6 \cdot 4}{0 \cdot 14}=45 \cdot 71$

Calculation of eq. wt.

$\frac{1 \cdot 878}{1 \cdot 0}=\frac{E+80}{E+35 \cdot 5} \quad$ where $E$ is the eq. wt. of metal.

$\text { or, } \quad \begin{aligned}1 \cdot 878(E+35 \cdot 5) &=E+80, \quad 1 \cdot 878 E+66 \cdot 67=E+80 \\E(1 \cdot 878-1 \cdot 0) &=80-66 \cdot 67\end{aligned}$

or,

$0 \cdot 878 E=13 \cdot 33, \quad \therefore E=\frac{13 \cdot 33}{0 \cdot 878}=15 \cdot 18=\text { eq. wt. of metal }$

Now, valency $=\frac{\approx \text { at. wt. }}{\text { eq. wt. }}=\frac{45 \cdot 71}{15 \cdot 18}=3 \cdot 01 \approx 3$ and exact atomic weight $=15 \cdot 18 \times 3=45 \cdot 54$

If the symbol of metal be $M$, formula of its bromide $=M B r_{3}$

$\therefore$ Molccular weight of $M B r_{3}=$ at. wt. of $M+3 \times$ at. wt. of $B r$

$=45 \cdot 54+80 \times 3=45 \cdot 54+240= 2 8 5 \cdot 5 4 \text {. }$