
The specific heat of a bivalent metal is 0.16. The approximate equivalent mass of the metal will be:
A.$40$
B.$20$
C.$80$
D.$10$
Answer
469.5k+ views
Hint: Equivalent mass of a metal is calculated by dividing its atomic mass by its valence. Atomic mass can be found out from specific heat using Dulong-Petit law.
Complete step by step answer:
Dulong-Petit law states that atoms of all elements have the same heat capacity. Therefore, the specific heat will be inversely proportional to the atomic mass of elements. Hence, the product of atomic mass and specific heat of an element in solid state will be constant, and its value is approximately equal to $6.4$.
i.e. atomic mass $ \times $ specific heat = $6.4$
We can find out the atomic mass of the element using the above equation.
Hence,
Atomic mass = $\dfrac{{6.4}}{\text{specific heat}}$
Given that, specific heat is equal to$0.16$. Let us substitute this value to the above equation.
Atomic mass = $\dfrac{{6.4}}{{0.16}} = 40$
i.e. the atomic mass of the element is $40$.
Given that the metal is bivalent. Therefore, its valency is $2$. We can find out the equivalent mass by dividing the atomic mass by $2$.
Equivalent mass = $\dfrac{\text{atomic mass}}{2}$
Equivalent mass = $\dfrac{{40}}{2} = 20$
Hence the approximate equivalent mass of the metal is $20$.
Additional information-
Dulong-Petit law can be applied to only elements which are in solid state. It cannot be applied to lighter elements having high boiling point. These are the limitations of the law. The product of atomic mass and specific heat of an element in solid state is called atomic heat of element.
Note:
Dulong-Petit law gives only the approximate atomic mass. Hence we can calculate only the approximate equivalent mass using this law.
Complete step by step answer:
Dulong-Petit law states that atoms of all elements have the same heat capacity. Therefore, the specific heat will be inversely proportional to the atomic mass of elements. Hence, the product of atomic mass and specific heat of an element in solid state will be constant, and its value is approximately equal to $6.4$.
i.e. atomic mass $ \times $ specific heat = $6.4$
We can find out the atomic mass of the element using the above equation.
Hence,
Atomic mass = $\dfrac{{6.4}}{\text{specific heat}}$
Given that, specific heat is equal to$0.16$. Let us substitute this value to the above equation.
Atomic mass = $\dfrac{{6.4}}{{0.16}} = 40$
i.e. the atomic mass of the element is $40$.
Given that the metal is bivalent. Therefore, its valency is $2$. We can find out the equivalent mass by dividing the atomic mass by $2$.
Equivalent mass = $\dfrac{\text{atomic mass}}{2}$
Equivalent mass = $\dfrac{{40}}{2} = 20$
Hence the approximate equivalent mass of the metal is $20$.
Additional information-
Dulong-Petit law can be applied to only elements which are in solid state. It cannot be applied to lighter elements having high boiling point. These are the limitations of the law. The product of atomic mass and specific heat of an element in solid state is called atomic heat of element.
Note:
Dulong-Petit law gives only the approximate atomic mass. Hence we can calculate only the approximate equivalent mass using this law.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Trending doubts
State and prove Bernoullis theorem class 11 physics CBSE

Raindrops are spherical because of A Gravitational class 11 physics CBSE

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

Write the differences between monocot plants and dicot class 11 biology CBSE

Why is steel more elastic than rubber class 11 physics CBSE

Explain why a There is no atmosphere on the moon b class 11 physics CBSE
