
Identify the correct statement
A. Gypsum contains a lower percentage of plaster of calcium than plaster of Paris
B. Gypsum is obtained by heating plaster of Paris
C. Plaster of Paris can be obtained by hydration of gypsum
D. Plaster of Paris is obtained by partial oxidation of gypsum
Answer
219.9k+ views
Hint: Gypsum has the chemical formula \[{\rm{CaS}}{{\rm{O}}_{\rm{4}}}{\rm{.2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\]. When heated at 393 K, it gives the compound calcium sulphate . When it is heated in air, water gets removed and converted initially to calcium sulphate hemihydrate or plaster and, if heated, more conversion to anhydrous calcium sulphate.
Complete Step by Step Answer:
Gypsum is widely excavated and also used as a fertiliser. It is the major unit in several structures of plaster, blackboard or sidewalk chalk, and drywall. The arrangement of gypsum has tiers of calcium and sulphate ions that are tightly bound together. These layers are linked by sheets of anion water molecules through weaker hydrogen bonding. This compound, when heated at 393 K, the water of crystallisation is lost and forms calcium sulphate 's calcium hemihydrate. Hemihydrate of calcium sulphate is also known as plaster of Paris.
The reaction occurs as follows: \[2(CaS{O_4}.2{H_2}O)\overset{393K}{\rightarrow}2(CaS{O_4}).{H_2}O\]\[ + 3{H_2}O\]
The temperature must not rise above 473K, leading to a loss of whole water. Anhydrous salt will be left called dead burnt plaster having no plaster of Paris properties. Plaster of Paris is a white powder.
The formula of plaster of paris is \[2\left( {{\rm{CaS}}{{\rm{O}}_{\rm{4}}}} \right){\rm{.}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\].
By analysing the formula of gypsum and plaster of Paris we see that gypsum contains a lower percentage of plaster calcium than plaster of Paris.
So, option A is correct.
Note: Plaster of Paris is utilised for generating moulds for pottery, ceramics, etc. It is employed for preparing statues, models, and other ornamental entities. It is utilised in surgical bandages for fixing torn and fractured bones in the body by immobilising the affected part of the body that has the fracture or sprain.
Complete Step by Step Answer:
Gypsum is widely excavated and also used as a fertiliser. It is the major unit in several structures of plaster, blackboard or sidewalk chalk, and drywall. The arrangement of gypsum has tiers of calcium and sulphate ions that are tightly bound together. These layers are linked by sheets of anion water molecules through weaker hydrogen bonding. This compound, when heated at 393 K, the water of crystallisation is lost and forms calcium sulphate 's calcium hemihydrate. Hemihydrate of calcium sulphate is also known as plaster of Paris.
The reaction occurs as follows: \[2(CaS{O_4}.2{H_2}O)\overset{393K}{\rightarrow}2(CaS{O_4}).{H_2}O\]\[ + 3{H_2}O\]
The temperature must not rise above 473K, leading to a loss of whole water. Anhydrous salt will be left called dead burnt plaster having no plaster of Paris properties. Plaster of Paris is a white powder.
The formula of plaster of paris is \[2\left( {{\rm{CaS}}{{\rm{O}}_{\rm{4}}}} \right){\rm{.}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\].
By analysing the formula of gypsum and plaster of Paris we see that gypsum contains a lower percentage of plaster calcium than plaster of Paris.
So, option A is correct.
Note: Plaster of Paris is utilised for generating moulds for pottery, ceramics, etc. It is employed for preparing statues, models, and other ornamental entities. It is utilised in surgical bandages for fixing torn and fractured bones in the body by immobilising the affected part of the body that has the fracture or sprain.
Recently Updated Pages
Electricity and Magnetism Explained: Key Concepts & Applications

JEE Energetics Important Concepts and Tips for Exam Preparation

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

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

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

States of Matter Chapter For JEE Main Chemistry

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
NCERT Solutions For Class 11 Chemistry Chapter 7 Redox Reaction

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

Thermodynamics Class 11 Chemistry Chapter 5 CBSE Notes - 2025-26

NCERT Solutions ForClass 11 Chemistry Chapter Chapter 5 Thermodynamics

Hydrocarbons Class 11 Chemistry Chapter 9 CBSE Notes - 2025-26

Equilibrium Class 11 Chemistry Chapter 6 CBSE Notes - 2025-26

