
In paraffins, with the increasing molecular weight, it is found that.
(a) Freezing point decreases
(b) Boiling point decreases
(c) Boiling point increases
(d) Vapour pressure decreases
Answer
160.8k+ views
Hint: When the intermolecular attraction between the molecules of a substance is suppressed by the energy of thermal motion. This can be easily understood that the kinetic energy is directionally proportional to the mass and velocity. Generally, the small molecules (with lower molecular weight) possess a lower boiling point than the large molecules (with high molecular weight). But there are some exceptions in which small molecules possess very high boiling points (such as water) compared to other small molecules due to the presence of very strong intermolecular forces.
Complete Step by Step Answer:
Paraffin is a saturated hydrocarbon that is composed of carbon and hydrogen as the major constituent elements.
Paraffin is represented by a \[{C_n}{H_{2n + 2}}\] general formula, Here C represents a carbon atom, H represents a hydrogen atom and n is an integer.
The main source of paraffin is petroleum.
Due to the small electronegativity difference between carbon and hydrogen, the paraffin is considered non-polar and they have the tendency to dissolve in non-polar solvents.
The boiling point of paraffin increases with the rise in molecular weight.
Paraffins with high molecular weight also possess a large surface area. This large surface area of paraffin causes more intermolecular interaction (Van Der Waals forces) between the molecules and as a result boiling point increases.
Therefore, from the above discussion, it is quite clear that option (c) will be the correct answer.
Note: Alkanes have inert and non-polar behaviour with very less reactivity. The availability of methane is very low and it is counted as the main component of greenhouse gases. The branched paraffin has a very slower degradation process than the unbranched paraffin.
Complete Step by Step Answer:
Paraffin is a saturated hydrocarbon that is composed of carbon and hydrogen as the major constituent elements.
Paraffin is represented by a \[{C_n}{H_{2n + 2}}\] general formula, Here C represents a carbon atom, H represents a hydrogen atom and n is an integer.
The main source of paraffin is petroleum.
Due to the small electronegativity difference between carbon and hydrogen, the paraffin is considered non-polar and they have the tendency to dissolve in non-polar solvents.
The boiling point of paraffin increases with the rise in molecular weight.
Paraffins with high molecular weight also possess a large surface area. This large surface area of paraffin causes more intermolecular interaction (Van Der Waals forces) between the molecules and as a result boiling point increases.
Therefore, from the above discussion, it is quite clear that option (c) will be the correct answer.
Note: Alkanes have inert and non-polar behaviour with very less reactivity. The availability of methane is very low and it is counted as the main component of greenhouse gases. The branched paraffin has a very slower degradation process than the unbranched paraffin.
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 Main Mock Test Series Class 12 Chemistry for FREE

Classification of Drugs Based on Pharmacological Effect, Drug Action

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics 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

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

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2026 Syllabus PDF - Download Paper 1 and 2 Syllabus by NTA

JEE Main Eligibility Criteria 2025

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

NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

NCERT Solutions for Class 12 Chemistry Chapter 2 Electrochemistry

NCERT Solutions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes

Solutions Class 12 Notes: CBSE Chemistry Chapter 1
