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29.2 % (w/w) HCl stock solution has density of 1.25 g mL1. 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
Correct answer is '8'. Can you explain this answer?
Verified Answer
29.2 % (w/w) HCl stock solution has density of 1.25 g mL1. The molecul...
Let mass of the stock solution = 100g
Mass of HCl in 100 g of 29.2 % (w/w) HCl stock solution = 29.2 g
Volume of the stock solution = 100g/1.25 g mL1 = 80 g
Number of moles of HCl in stock solution = 29.2/36.5 = 0.8
Molarity of the stock solution = (0.82/80)×1000 = 10 M
Using,
M1V1 = M2V2
10× V1 =0.4×200
or
V1 = 8mL
Hence, the volume required is 8 mL.
Most Upvoted Answer
29.2 % (w/w) HCl stock solution has density of 1.25 g mL1. The molecul...
To determine 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, let's calculate the number of moles of HCl needed for the 0.4 M solution:
Molarity = (moles of HCl) / (0.200 L)
0.4 M = (moles of HCl) / 0.200 L
Rearranging the equation, we find:
moles of HCl = 0.4 M × 0.200 L
moles of HCl = 0.08 mol
Now, let's calculate the grams of HCl needed to make 0.08 mol of HCl:
grams of HCl = moles of HCl × molecular weight of HCl
grams of HCl = 0.08 mol × 36.5 g/mol
grams of HCl = 2.92 g
Since we have a 29.2% (w/w) HCl stock solution with a density of 1.25 g/mL, we can calculate the volume of the stock solution needed to obtain 2.92 g of HCl:
volume of stock solution = (grams of HCl) / (density of stock solution)
volume of stock solution = 2.92 g / 1.25 g/mL
volume of stock solution = 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. However, the answer is given as '8'. Since the calculation shows a different value, it is possible that there might be an error in the given answer.
Molarity (M) = (moles of solute) / (volume of solution in liters)
First, let's calculate the number of moles of HCl needed for the 0.4 M solution:
Molarity = (moles of HCl) / (0.200 L)
0.4 M = (moles of HCl) / 0.200 L
Rearranging the equation, we find:
moles of HCl = 0.4 M × 0.200 L
moles of HCl = 0.08 mol
Now, let's calculate the grams of HCl needed to make 0.08 mol of HCl:
grams of HCl = moles of HCl × molecular weight of HCl
grams of HCl = 0.08 mol × 36.5 g/mol
grams of HCl = 2.92 g
Since we have a 29.2% (w/w) HCl stock solution with a density of 1.25 g/mL, we can calculate the volume of the stock solution needed to obtain 2.92 g of HCl:
volume of stock solution = (grams of HCl) / (density of stock solution)
volume of stock solution = 2.92 g / 1.25 g/mL
volume of stock solution = 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. However, the answer is given as '8'. Since the calculation shows a different value, it is possible that there might be an error in the given answer.
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29.2 % (w/w) HCl stock solution has density of 1.25 g mL1. 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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29.2 % (w/w) HCl stock solution has density of 1.25 g mL1. 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 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 29.2 % (w/w) HCl stock solution has density of 1.25 g mL1. 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 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 29.2 % (w/w) HCl stock solution has density of 1.25 g mL1. 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?.
29.2 % (w/w) HCl stock solution has density of 1.25 g mL1. 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 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 29.2 % (w/w) HCl stock solution has density of 1.25 g mL1. 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 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 29.2 % (w/w) HCl stock solution has density of 1.25 g mL1. 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 29.2 % (w/w) HCl stock solution has density of 1.25 g mL1. 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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