Question

The chloride of a solid metallic element contains by mass $57.89{}^0/{}_0$ of the element. The specific heat of the element is$0.0324\text{ de}{\text{g}}^{\text{-1}}{\text{g}}^{\text{-1}}$. Calculate the exact atomic mass of the element.
A) $$195.2$$
B) $$195$$
C) $$195.9$$
D)$$194$$

Answer 1

Hint: The atomic mass of the elements can be obtained by calculating the equivalent weight of the metal. The approximate weight of the metal can be calculated by the Dulong and Petit law. According to which the approximate weight of the element is$\text{Approx}\text{.weight of metal}\times \text{Specific heat of metal=6}\text{.4}$. Further the actual or exact weight of the element is found out through the relation is:
$$\text{Approx}\text{. atomic wt}\text{.}=\text{Equivalent wt}\text{.}\times \text{Valency of metal}$$

Complete step by step solution:
Here, we are provided with the following data,
${}^0/{}_0$ Of the metal which is provided in the question$=57.89{}^0/{}_0$
The specific heat of the element is $0.0324\text{ de}{\text{g}}^{\text{-1}}{\text{g}}^{\text{-1}}$
The mass percentage is a method to represent the concentration of any element in the compound of a mixture of the component. Mass percentage is calculated as the ratio of the mass of a component and the total mass of the mixture multiplied by the$100{}^0/{}_0$.
Since we know that the mass percentage for a mixture is always equal to$100{}^0/{}_0$.
Therefore the ${}^0/{}_0$ of the chlorine can be calculated as
$\begin{array}{rl}& =(100-57.89){}^0/{}_0\\ & =42.11{}^0/{}_0\end{array}$
We have to find out the atomic weight of the metal.
Let’s first calculate the equivalent weight of the metal
$\text{Equivalent weight of metal=}{\displaystyle \frac{{}^{\text{0}}/{}_{\text{0}}\text{metal}}{{}^{\text{0}}/{}_{\text{0}}\text{Chlorine}}}\times \text{ Atomic weight of Cl}$
Substitute the values of ${}^{\text{0}}/{}_{\text{0}}\text{metal}$and ${}^{\text{0}}/{}_{\text{0}}\text{Chlorine}$in the equation. We get,
$\text{Equivalent weight of metal=}{\displaystyle \frac{57.89}{\text{42}\text{.11}}}\times \text{ 35}\text{.5=48}\text{.8 g/mol}$
(Since the atomic weight of chlorine is$35.5\text{ g/mol}$)
We are given the specific heat of metal as$0.0324\text{ de}{\text{g}}^{\text{-1}}{\text{g}}^{\text{-1}}$.
According to the Dulong and Petit’s law, it is found that the product of the atomic mass of the metal and that of the specific heat of element is approximately equal to the$$\text{6}\text{.4 cal/mol}$$. Using the Dulong and Petit law we can relate the specific heat of the given element concerning the approximate weight of the element found out.
The approximate weight of the metal is determined as follows:
$\text{Approximate weight of metal=}{\displaystyle \frac{\text{6}\text{.4}}{\text{Specific heat of metal}}}$
Let's substitute the value for the specific heat.
$\text{Approximate weight of metal=}{\displaystyle \frac{\text{6}\text{.4}}{0.0324}}=197.53\text{ g}$
We know that the valency of metal is directly related to the ratio of approximate atomic weight and the equivalent weight of metal. The relation is
$\text{Valency of metal=}{\displaystyle \frac{\text{approximate atomic weight}}{\text{Equivalent weight}}}$
Substitute the values for equivalent weight and approx.atomic weight to find the valency of metal.
$\text{Valency of metal=}{\displaystyle \frac{\text{197}\text{.53}}{\text{48}\text{.8}}}$
$\text{Valency of metal=4}\text{.04}\simeq \text{4}$
Therefore, the metal has a valency of 4.
Let's find out the exact atomic weight of the metal. Since we know the valency and equivalent weight of metal.
$\text{Exact atomic weight of metal=Equivalent weight of metal}\times \text{Valency of metal}$
$\text{Exact atomic weight of metal=48}\text{.8}\times \text{4}$
Or $\text{Exact atomic weight of metal=}195.2$
Therefore the exact atomic weight of metal is equal to the$195.2$.

Hence, (A) is the correct option.

Note: Dulong and Petit's law applies to solid elements only. According to the law, the atomic heat of elements is equal to 6.4.