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4.5 gram of aluminium(atomic mass 27 amu) is deposited at cathode from Al3 solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from H ions in solution by the same quantity of electric charge will be a)44.8L b)22.4L c)11.2L d)5.6L?
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4.5 gram of aluminium(atomic mass 27 amu) is deposited at cathode from...
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4.5 gram of aluminium(atomic mass 27 amu) is deposited at cathode from...
Calculation of moles of aluminum:
Given: Mass of aluminum = 4.5 grams, Atomic mass of aluminum = 27 amu

To calculate the moles of aluminum, we can use the formula:
Moles = Mass / Atomic mass

Moles of aluminum = 4.5 grams / 27 amu
Moles of aluminum = 0.1667 moles

Calculation of moles of Al3+ ions:
Since each Al3+ ion has a charge of +3, the moles of Al3+ ions will be three times the moles of aluminum.

Moles of Al3+ ions = 0.1667 moles * 3
Moles of Al3+ ions = 0.5 moles

Calculation of moles of hydrogen:
Since the electric charge is the same for both aluminum and hydrogen, the moles of hydrogen can be calculated using the moles of Al3+ ions.

Moles of hydrogen = Moles of Al3+ ions
Moles of hydrogen = 0.5 moles

Calculation of volume of hydrogen at STP:
Given: 1 mole of any gas occupies 22.4 L at STP

Volume of hydrogen = Moles of hydrogen * 22.4 L/mol
Volume of hydrogen = 0.5 moles * 22.4 L/mol
Volume of hydrogen = 11.2 L

Therefore, the volume of hydrogen produced at STP from H+ ions in solution by the same quantity of electric charge is 11.2 L (option c).
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4.5 gram of aluminium(atomic mass 27 amu) is deposited at cathode from Al3 solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from H ions in solution by the same quantity of electric charge will be a)44.8L b)22.4L c)11.2L d)5.6L?
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