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SECTION III
Q. No. 36 - 40 carry 4 marks each
The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive).
Q.
29.2 % (w/w) HCl stock solution has density of 1.25 g mL-1. The molecular weight of HCl is 36.5 g mol-1.
The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is
The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl is
Correct answer is '8'. Can you explain this answer?
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SECTION IIIQ. No. 36 - 40 carry 4 marks eachThe answer to each questio...
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SECTION IIIQ. No. 36 - 40 carry 4 marks eachThe answer to each questio...
To find the volume of the stock solution required to prepare a 200 mL solution of 0.4 M HCl, we can use the formula:
Molarity (M) = (moles of solute) / (volume of solution in liters)
First, we need to calculate the number of moles of HCl required in the 200 mL solution. The formula to calculate moles is:
moles = Molarity × volume (in liters)
Given that the molarity is 0.4 M and the volume is 200 mL, we convert the volume to liters:
volume = 200 mL × (1 L / 1000 mL) = 0.2 L
Now we can calculate the moles of HCl required:
moles = 0.4 M × 0.2 L = 0.08 moles
Next, we need to find the mass of HCl required in grams. The formula to calculate mass is:
mass = moles × molar mass
The molar mass of HCl is given as 36.5 g/mol. So:
mass = 0.08 moles × 36.5 g/mol = 2.92 g
Since the density of the stock solution is given as 1.25 g/mL, we can calculate the volume of the stock solution required:
volume of stock solution = mass / density
volume of stock solution = 2.92 g / 1.25 g/mL = 2.336 mL
Therefore, the volume of the stock solution required to prepare a 200 mL solution of 0.4 M HCl is approximately 2.336 mL.
Answer: 2.336 mL
Molarity (M) = (moles of solute) / (volume of solution in liters)
First, we need to calculate the number of moles of HCl required in the 200 mL solution. The formula to calculate moles is:
moles = Molarity × volume (in liters)
Given that the molarity is 0.4 M and the volume is 200 mL, we convert the volume to liters:
volume = 200 mL × (1 L / 1000 mL) = 0.2 L
Now we can calculate the moles of HCl required:
moles = 0.4 M × 0.2 L = 0.08 moles
Next, we need to find the mass of HCl required in grams. The formula to calculate mass is:
mass = moles × molar mass
The molar mass of HCl is given as 36.5 g/mol. So:
mass = 0.08 moles × 36.5 g/mol = 2.92 g
Since the density of the stock solution is given as 1.25 g/mL, we can calculate the volume of the stock solution required:
volume of stock solution = mass / density
volume of stock solution = 2.92 g / 1.25 g/mL = 2.336 mL
Therefore, the volume of the stock solution required to prepare a 200 mL solution of 0.4 M HCl is approximately 2.336 mL.
Answer: 2.336 mL
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SECTION IIIQ. No. 36 - 40 carry 4 marks eachThe answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive).Q.29.2 % (w/w) HCl stock solution has density of 1.25 g mL-1. The molecular weight of HCl is 36.5 g mol-1.The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl isCorrect answer is '8'. Can you explain this answer?
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SECTION IIIQ. No. 36 - 40 carry 4 marks eachThe answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive).Q.29.2 % (w/w) HCl stock solution has density of 1.25 g mL-1. The molecular weight of HCl is 36.5 g mol-1.The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl isCorrect answer is '8'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about SECTION IIIQ. No. 36 - 40 carry 4 marks eachThe answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive).Q.29.2 % (w/w) HCl stock solution has density of 1.25 g mL-1. The molecular weight of HCl is 36.5 g mol-1.The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl isCorrect answer is '8'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for SECTION IIIQ. No. 36 - 40 carry 4 marks eachThe answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive).Q.29.2 % (w/w) HCl stock solution has density of 1.25 g mL-1. The molecular weight of HCl is 36.5 g mol-1.The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl isCorrect answer is '8'. Can you explain this answer?.
SECTION IIIQ. No. 36 - 40 carry 4 marks eachThe answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive).Q.29.2 % (w/w) HCl stock solution has density of 1.25 g mL-1. The molecular weight of HCl is 36.5 g mol-1.The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl isCorrect answer is '8'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about SECTION IIIQ. No. 36 - 40 carry 4 marks eachThe answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive).Q.29.2 % (w/w) HCl stock solution has density of 1.25 g mL-1. The molecular weight of HCl is 36.5 g mol-1.The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl isCorrect answer is '8'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for SECTION IIIQ. No. 36 - 40 carry 4 marks eachThe answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive).Q.29.2 % (w/w) HCl stock solution has density of 1.25 g mL-1. The molecular weight of HCl is 36.5 g mol-1.The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl isCorrect answer is '8'. Can you explain this answer?.
Solutions for SECTION IIIQ. No. 36 - 40 carry 4 marks eachThe answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive).Q.29.2 % (w/w) HCl stock solution has density of 1.25 g mL-1. The molecular weight of HCl is 36.5 g mol-1.The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl isCorrect answer is '8'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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ample number of questions to practice SECTION IIIQ. No. 36 - 40 carry 4 marks eachThe answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive).Q.29.2 % (w/w) HCl stock solution has density of 1.25 g mL-1. The molecular weight of HCl is 36.5 g mol-1.The volume (mL) of stock solution required to prepare a 200 mL solution of 0.4 M HCl isCorrect answer is '8'. Can you explain this answer? tests, examples and also practice JEE tests.
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