
\[1.86\,g\] of aniline completely reacts to form acetanilide. \[10\% \] of the product is lost during purification. Amount of acetanilide after purification (in g) is ________\[ \times {10^{ - 2}}\].
Answer
163.8k+ views
Hint: There is \[1.86\,g\] of aniline is given. For solving this question, we have to calculate the molar mass of aniline and acetanilide. The structure of aniline is as follows:

Image: Structure of aniline
The structure of acetanilide is as follows:

Image: Structure of acetanilide
These structures are useful to understand the conversion of aniline to acetanilide.
Complete Step by Step Solution:
For calculating the amount of acetanilide obtained, we have to calculate the molar mass of aniline and acetanilide.
The chemical formula of aniline is \[{C_6}{H_7}N\] .
Let us calculate the molar mass of aniline as follows:
$ {\text{molar mass of aniline}} = 6\left( {12} \right) + 7\left( 1 \right) + 1\left( {14} \right) \\$
$ \Rightarrow {\text{molar mass of aniline}} = 93 \\ $
Therefore, the molar mass of aniline is 93.
The chemical formula of acetanilide is \[{C_8}{H_9}NO\] .
Let us calculate the molar mass of acetanilide as follows:
$ {\text{molar mass of acetanilide}} = 8\left( {12} \right) + 9\left( 1 \right) + 1\left( {14} \right) + 1\left( {16} \right) \\$
$ \Rightarrow {\text{molar mass of acetanilide}} = 135 \\ $
Therefore, the molar mass of acetanilide is 135.

Image: Chemical reaction for preparation of acetanilide
From the reaction stoichiometry, one-mole aniline produces one mole of acetanilide.
Now, by using molar mass we can say that,
\[93\,g\] aniline produces \[135\,g\] acetanilide.
Let us calculate that acetanilide produces by \[1.86\,g\] of aniline as follows:
$ {\text{amount of acetanilide}} = 1.86 \times \dfrac{{135}}{{93}} \\$
$ \Rightarrow {\text{amount of acetanilide}} = 2.7\,g \\ $
Therefore, there is \[2.7\,g\] of acetanilide produced.
But \[10\% \] of the product is lost during purification. Therefore, there is \[90\% \] product formed.
Let us calculate the actual product as follows:
$ {\text{actual amount of acetanilide}} = 2.7 \times \dfrac{{90}}{{100}} \\$
$ \Rightarrow {\text{actual amount of acetanilide}} = 2.43\,g \\$
$ \Rightarrow {\text{actual amount of acetanilide}} = 243 \times {10^{ - 2}}\,g \\ $
As a result, there is \[243 \times {10^{ - 2}}\,g\] of acetanilide produced in the reaction.
Therefore, correct answer is \[243\] .
Note: Acetanilide is a carcinogenic substance that can lead to cancer in people. During the fractional distillation of acetic anhydride, avoid inhaling the vapours. Continue to boil the mixture vigorously while stirring in the ice-cold water. Because of the amide group, acetanilide is a polar molecule. Antifebrin was the brand name for a drug that was once used to treat fever and headaches.

Image: Structure of aniline
The structure of acetanilide is as follows:

Image: Structure of acetanilide
These structures are useful to understand the conversion of aniline to acetanilide.
Complete Step by Step Solution:
For calculating the amount of acetanilide obtained, we have to calculate the molar mass of aniline and acetanilide.
The chemical formula of aniline is \[{C_6}{H_7}N\] .
Let us calculate the molar mass of aniline as follows:
$ {\text{molar mass of aniline}} = 6\left( {12} \right) + 7\left( 1 \right) + 1\left( {14} \right) \\$
$ \Rightarrow {\text{molar mass of aniline}} = 93 \\ $
Therefore, the molar mass of aniline is 93.
The chemical formula of acetanilide is \[{C_8}{H_9}NO\] .
Let us calculate the molar mass of acetanilide as follows:
$ {\text{molar mass of acetanilide}} = 8\left( {12} \right) + 9\left( 1 \right) + 1\left( {14} \right) + 1\left( {16} \right) \\$
$ \Rightarrow {\text{molar mass of acetanilide}} = 135 \\ $
Therefore, the molar mass of acetanilide is 135.

Image: Chemical reaction for preparation of acetanilide
From the reaction stoichiometry, one-mole aniline produces one mole of acetanilide.
Now, by using molar mass we can say that,
\[93\,g\] aniline produces \[135\,g\] acetanilide.
Let us calculate that acetanilide produces by \[1.86\,g\] of aniline as follows:
$ {\text{amount of acetanilide}} = 1.86 \times \dfrac{{135}}{{93}} \\$
$ \Rightarrow {\text{amount of acetanilide}} = 2.7\,g \\ $
Therefore, there is \[2.7\,g\] of acetanilide produced.
But \[10\% \] of the product is lost during purification. Therefore, there is \[90\% \] product formed.
Let us calculate the actual product as follows:
$ {\text{actual amount of acetanilide}} = 2.7 \times \dfrac{{90}}{{100}} \\$
$ \Rightarrow {\text{actual amount of acetanilide}} = 2.43\,g \\$
$ \Rightarrow {\text{actual amount of acetanilide}} = 243 \times {10^{ - 2}}\,g \\ $
As a result, there is \[243 \times {10^{ - 2}}\,g\] of acetanilide produced in the reaction.
Therefore, correct answer is \[243\] .
Note: Acetanilide is a carcinogenic substance that can lead to cancer in people. During the fractional distillation of acetic anhydride, avoid inhaling the vapours. Continue to boil the mixture vigorously while stirring in the ice-cold water. Because of the amide group, acetanilide is a polar molecule. Antifebrin was the brand name for a drug that was once used to treat fever and headaches.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

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

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

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

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

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

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

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

Instantaneous Velocity - Formula based Examples for JEE

Thermodynamics Class 11 Notes: CBSE Chapter 5
