NEET Exam > NEET Questions > 1g of pure calcium carbonate was found to req...
Start Learning for Free
1g of pure calcium carbonate was found to require 50ml dil. HCL for complete reaction.The strength if the HCL solution is given by. a) 4 N b ) 2N c) 0.4 N d) 0.2 N?
Verified Answer
1g of pure calcium carbonate was found to require 50ml dil. HCL for co...
This question is part of UPSC exam. View all NEET courses
Most Upvoted Answer
1g of pure calcium carbonate was found to require 50ml dil. HCL for co...
Calculation of the Strength of HCL Solution
Given:
Mass of pure calcium carbonate = 1g
Volume of dilute HCL solution required for complete reaction = 50ml
We can determine the strength of the HCL solution by using the concept of normality (N). Normality is defined as the number of equivalents of solute present in one liter of solution.
To calculate the normality of the HCL solution, we need to find the number of equivalents of HCL present in the volume used for the reaction. The equation for the reaction between calcium carbonate and hydrochloric acid is as follows:
CaCO3 + 2HCl -> CaCl2 + CO2 + H2O
From the balanced chemical equation, we can see that 1 mole of calcium carbonate reacts with 2 moles of hydrochloric acid. Since we know the mass of calcium carbonate used, we can calculate the number of moles of calcium carbonate.
Molar mass of calcium carbonate (CaCO3) = 40.08g/mol + 12.01g/mol + (3 * 16.00g/mol) = 100.09g/mol
Number of moles of calcium carbonate = Mass / Molar mass = 1g / 100.09g/mol = 0.00999mol ≈ 0.01mol
Since 1 mole of calcium carbonate reacts with 2 moles of hydrochloric acid, the number of moles of hydrochloric acid required for the reaction is twice the number of moles of calcium carbonate.
Number of moles of HCl = 2 * 0.01mol = 0.02mol
Now, we can calculate the normality of the HCL solution using the formula:
Normality (N) = Number of equivalents / Volume of solution (in liters)
To calculate the number of equivalents, we need to know the equivalent weight of HCL. The equivalent weight of an acid is the molar mass divided by the number of acidic hydrogen atoms.
Molar mass of HCL = 1.01g/mol + 35.45g/mol = 36.46g/mol
The number of acidic hydrogen atoms in HCL is 1.
Equivalent weight of HCL = Molar mass / Number of acidic hydrogen atoms = 36.46g/mol / 1 = 36.46g/mol
Number of equivalents of HCL = Number of moles of HCL * Equivalent weight of HCL
= 0.02mol * 36.46g/mol = 0.7292g
Now, we can calculate the normality of the HCL solution:
Normality (N) = Number of equivalents / Volume of solution (in liters)
= 0.7292g / 0.05L = 14.584 N
Therefore, the strength of the HCL solution is approximately 14.584 N.
Now, let's match it with the given options:
a) 4 N
b) 2 N
c) 0.4 N
d) 0.2 N
The calculated normality is much higher than any of the given options. Therefore, none of the provided options match the strength of the HCL solution.
Given:
Mass of pure calcium carbonate = 1g
Volume of dilute HCL solution required for complete reaction = 50ml
We can determine the strength of the HCL solution by using the concept of normality (N). Normality is defined as the number of equivalents of solute present in one liter of solution.
To calculate the normality of the HCL solution, we need to find the number of equivalents of HCL present in the volume used for the reaction. The equation for the reaction between calcium carbonate and hydrochloric acid is as follows:
CaCO3 + 2HCl -> CaCl2 + CO2 + H2O
From the balanced chemical equation, we can see that 1 mole of calcium carbonate reacts with 2 moles of hydrochloric acid. Since we know the mass of calcium carbonate used, we can calculate the number of moles of calcium carbonate.
Molar mass of calcium carbonate (CaCO3) = 40.08g/mol + 12.01g/mol + (3 * 16.00g/mol) = 100.09g/mol
Number of moles of calcium carbonate = Mass / Molar mass = 1g / 100.09g/mol = 0.00999mol ≈ 0.01mol
Since 1 mole of calcium carbonate reacts with 2 moles of hydrochloric acid, the number of moles of hydrochloric acid required for the reaction is twice the number of moles of calcium carbonate.
Number of moles of HCl = 2 * 0.01mol = 0.02mol
Now, we can calculate the normality of the HCL solution using the formula:
Normality (N) = Number of equivalents / Volume of solution (in liters)
To calculate the number of equivalents, we need to know the equivalent weight of HCL. The equivalent weight of an acid is the molar mass divided by the number of acidic hydrogen atoms.
Molar mass of HCL = 1.01g/mol + 35.45g/mol = 36.46g/mol
The number of acidic hydrogen atoms in HCL is 1.
Equivalent weight of HCL = Molar mass / Number of acidic hydrogen atoms = 36.46g/mol / 1 = 36.46g/mol
Number of equivalents of HCL = Number of moles of HCL * Equivalent weight of HCL
= 0.02mol * 36.46g/mol = 0.7292g
Now, we can calculate the normality of the HCL solution:
Normality (N) = Number of equivalents / Volume of solution (in liters)
= 0.7292g / 0.05L = 14.584 N
Therefore, the strength of the HCL solution is approximately 14.584 N.
Now, let's match it with the given options:
a) 4 N
b) 2 N
c) 0.4 N
d) 0.2 N
The calculated normality is much higher than any of the given options. Therefore, none of the provided options match the strength of the HCL solution.
Attention NEET Students!
To make sure you are not studying endlessly, EduRev has designed NEET study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in NEET.
|
|
Explore Courses for NEET exam
|
|
Similar NEET Doubts
Top Courses for NEETView all
1g of pure calcium carbonate was found to require 50ml dil. HCL for complete reaction.The strength if the HCL solution is given by. a) 4 N b ) 2N c) 0.4 N d) 0.2 N?
Question Description
1g of pure calcium carbonate was found to require 50ml dil. HCL for complete reaction.The strength if the HCL solution is given by. a) 4 N b ) 2N c) 0.4 N d) 0.2 N? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 1g of pure calcium carbonate was found to require 50ml dil. HCL for complete reaction.The strength if the HCL solution is given by. a) 4 N b ) 2N c) 0.4 N d) 0.2 N? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1g of pure calcium carbonate was found to require 50ml dil. HCL for complete reaction.The strength if the HCL solution is given by. a) 4 N b ) 2N c) 0.4 N d) 0.2 N?.
1g of pure calcium carbonate was found to require 50ml dil. HCL for complete reaction.The strength if the HCL solution is given by. a) 4 N b ) 2N c) 0.4 N d) 0.2 N? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 1g of pure calcium carbonate was found to require 50ml dil. HCL for complete reaction.The strength if the HCL solution is given by. a) 4 N b ) 2N c) 0.4 N d) 0.2 N? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1g of pure calcium carbonate was found to require 50ml dil. HCL for complete reaction.The strength if the HCL solution is given by. a) 4 N b ) 2N c) 0.4 N d) 0.2 N?.
Solutions for 1g of pure calcium carbonate was found to require 50ml dil. HCL for complete reaction.The strength if the HCL solution is given by. a) 4 N b ) 2N c) 0.4 N d) 0.2 N? in English & in Hindi are available as part of our courses for NEET.
Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free.
Here you can find the meaning of 1g of pure calcium carbonate was found to require 50ml dil. HCL for complete reaction.The strength if the HCL solution is given by. a) 4 N b ) 2N c) 0.4 N d) 0.2 N? defined & explained in the simplest way possible. Besides giving the explanation of
1g of pure calcium carbonate was found to require 50ml dil. HCL for complete reaction.The strength if the HCL solution is given by. a) 4 N b ) 2N c) 0.4 N d) 0.2 N?, a detailed solution for 1g of pure calcium carbonate was found to require 50ml dil. HCL for complete reaction.The strength if the HCL solution is given by. a) 4 N b ) 2N c) 0.4 N d) 0.2 N? has been provided alongside types of 1g of pure calcium carbonate was found to require 50ml dil. HCL for complete reaction.The strength if the HCL solution is given by. a) 4 N b ) 2N c) 0.4 N d) 0.2 N? theory, EduRev gives you an
ample number of questions to practice 1g of pure calcium carbonate was found to require 50ml dil. HCL for complete reaction.The strength if the HCL solution is given by. a) 4 N b ) 2N c) 0.4 N d) 0.2 N? tests, examples and also practice NEET tests.
|
|
Explore Courses for NEET exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup