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A mixture of Al and Zn weighing 1.67 g was completely dissolved in acid, and the evolved 1.69 litres of hydrogen gas was measured at 273 K and one atmosphere pressure. The respective masses of Al and Zn in the original mixture were
  • a)
    1.25 g, 0.42 g
  • b)
    0.42 g, 1.25 g
  • c)
    1.25 g, 4.2 g
  • d)
    12.5 g, 4.2 g
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A mixture of Al and Zn weighing 1.67 g was completely dissolved in ac...
Let 'a' g and 'b 'g be the weights of Al and Zn in the mixture.
a + b = 1.67 … (1)
Meq of Al + Meq of Zn = Meq of H2
From equations (1) and (2)
a = 1.25 g Al
b = 0.42 g Zn
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Community Answer
A mixture of Al and Zn weighing 1.67 g was completely dissolved in ac...
Given:
- Total weight of Al and Zn mixture = 1.67 g
- Volume of hydrogen gas evolved = 1.69 L
- Temperature (T) = 273 K
- Pressure (P) = 1 atm

Calculations:
- Calculate the number of moles of hydrogen gas evolved using the ideal gas equation: PV = nRT
- n = (PV) / (RT)
- n = (1 atm * 1.69 L) / (0.0821 L.atm/mol.K * 273 K)
- n = 0.088 moles
- Assume the weights of Al and Zn in the mixture as x g and (1.67-x) g respectively
- Calculate the number of moles of Al and Zn using their respective molar masses
- Moles of Al = x / 27 g/mol
- Moles of Zn = (1.67 - x) / 65 g/mol
- From the chemical reaction:
2Al + 3ZnCl2 → 3Zn + 2AlCl3 + 3H2
- It can be observed that 2 moles of Al produce 3 moles of H2
- Therefore, the moles of Al in the mixture is related to the moles of H2 evolved as:
2 * (moles of Al) = 3 * 0.088
- Solving for x, we get x = 1.25 g
Therefore, the respective masses of Al and Zn in the original mixture were 1.25 g and 0.42 g, which corresponds to option 'A'.
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A mixture of Al and Zn weighing 1.67 g was completely dissolved in acid, and the evolved 1.69 litres of hydrogen gas was measured at 273 K and one atmosphere pressure. The respective masses of Al and Zn in the original mixture werea)1.25 g, 0.42 gb)0.42 g, 1.25 gc)1.25 g, 4.2 gd)12.5 g, 4.2 gCorrect answer is option 'A'. Can you explain this answer?
Question Description
A mixture of Al and Zn weighing 1.67 g was completely dissolved in acid, and the evolved 1.69 litres of hydrogen gas was measured at 273 K and one atmosphere pressure. The respective masses of Al and Zn in the original mixture werea)1.25 g, 0.42 gb)0.42 g, 1.25 gc)1.25 g, 4.2 gd)12.5 g, 4.2 gCorrect answer is option 'A'. 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 A mixture of Al and Zn weighing 1.67 g was completely dissolved in acid, and the evolved 1.69 litres of hydrogen gas was measured at 273 K and one atmosphere pressure. The respective masses of Al and Zn in the original mixture werea)1.25 g, 0.42 gb)0.42 g, 1.25 gc)1.25 g, 4.2 gd)12.5 g, 4.2 gCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A mixture of Al and Zn weighing 1.67 g was completely dissolved in acid, and the evolved 1.69 litres of hydrogen gas was measured at 273 K and one atmosphere pressure. The respective masses of Al and Zn in the original mixture werea)1.25 g, 0.42 gb)0.42 g, 1.25 gc)1.25 g, 4.2 gd)12.5 g, 4.2 gCorrect answer is option 'A'. Can you explain this answer?.
Solutions for A mixture of Al and Zn weighing 1.67 g was completely dissolved in acid, and the evolved 1.69 litres of hydrogen gas was measured at 273 K and one atmosphere pressure. The respective masses of Al and Zn in the original mixture werea)1.25 g, 0.42 gb)0.42 g, 1.25 gc)1.25 g, 4.2 gd)12.5 g, 4.2 gCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
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