
Ammonium carbamate when heated to $200{}^\circ C$ gives a mixture of \[N{{H}_{3}}\] and \[C{{O}_{2}}\] vapour with a density of 13. What is the degree of dissociation of ammonium carbamate?
A. \[\dfrac{3}{2}\]
B. $\dfrac{3}{4}$
C. $2$
D. $1$
Answer
163.5k+ views
Hint: The percentage of an electrolyte's total moles that disintegrate into its ions when equilibrium is reached is referred to as the electrolyte's degree of dissociation. It’s denoted by $\alpha $.
Formula used: $D=\dfrac{m.wt.}{2}$ where D is the density of ammonium carbamate and m.wt. is the molecular mass of ammonium carbamate. $\alpha =\dfrac{D-d}{(n-1)d}$ where D is the density of ammonium carbamate and d is the given vapour density. n is the number of products.
Complete Step by Step Solution:
Degree of dissociation is defined as the percentage of an electrolyte’s total moles that disintegrates into its ions when equilibrium is reached.
According to given question, ammonium carbamate is disintegrated to give a mixture of \[N{{H}_{3}}\] and \[C{{O}_{2}}\].
Molecular formula of Ammonium carbamate is $N{{H}_{2}}COON{{H}_{4}}$
Molecular mass of Ammonium carbamate= 78
When it’s heated at equilibrium the reaction is given as:
$N{{H}_{2}}COON{{H}_{{{4}_{(s)}}}}2N{{H}_{{{3}_{(g)}}}}+C{{O}_{{{2}_{(g)}}}}$
Now using the formula:
$D=\dfrac{m.wt.}{2}$, putting the values of m.wt= 78
Therefore, $D=\dfrac{78}{2}=39$
Now, we need to find $\alpha $using the formula $\alpha =\dfrac{D-d}{(n-1)d}$
We can see from the equilibrium reaction that $n=3$ and $d=13$, which is given in the question.
Now putting the appropriate values:
$\alpha =\dfrac{39-13}{(3-1)\times 13}$
Solving it further, we get
$\alpha =\dfrac{26}{26}=1$
Hence, the correct option is D. $1$
Note: You should know the basics of equilibrium and how to write and balance a chemical reaction at equilibrium. You must remember the basics of degree of dissociation and the formulas related to it. The different formulas used in the solution must be remembered.
Formula used: $D=\dfrac{m.wt.}{2}$ where D is the density of ammonium carbamate and m.wt. is the molecular mass of ammonium carbamate. $\alpha =\dfrac{D-d}{(n-1)d}$ where D is the density of ammonium carbamate and d is the given vapour density. n is the number of products.
Complete Step by Step Solution:
Degree of dissociation is defined as the percentage of an electrolyte’s total moles that disintegrates into its ions when equilibrium is reached.
According to given question, ammonium carbamate is disintegrated to give a mixture of \[N{{H}_{3}}\] and \[C{{O}_{2}}\].
Molecular formula of Ammonium carbamate is $N{{H}_{2}}COON{{H}_{4}}$
Molecular mass of Ammonium carbamate= 78
When it’s heated at equilibrium the reaction is given as:
$N{{H}_{2}}COON{{H}_{{{4}_{(s)}}}}2N{{H}_{{{3}_{(g)}}}}+C{{O}_{{{2}_{(g)}}}}$
Now using the formula:
$D=\dfrac{m.wt.}{2}$, putting the values of m.wt= 78
Therefore, $D=\dfrac{78}{2}=39$
Now, we need to find $\alpha $using the formula $\alpha =\dfrac{D-d}{(n-1)d}$
We can see from the equilibrium reaction that $n=3$ and $d=13$, which is given in the question.
Now putting the appropriate values:
$\alpha =\dfrac{39-13}{(3-1)\times 13}$
Solving it further, we get
$\alpha =\dfrac{26}{26}=1$
Hence, the correct option is D. $1$
Note: You should know the basics of equilibrium and how to write and balance a chemical reaction at equilibrium. You must remember the basics of degree of dissociation and the formulas related to it. The different formulas used in the solution must be remembered.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Types of Solutions

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

NEET 2025 – Every New Update You Need to Know

Verb Forms Guide: V1, V2, V3, V4, V5 Explained
