$4.5 g$ of aluminum (at. mass 27amu) is deposited at cathode from $A l^{3 +}$ 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

22.4 L

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44.8 L

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5.6 L

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11.2 L

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Solution

The correct option is C 5.6 L
$E q o f A l = e q o f H_{2}$
$\frac{4.5}{\frac{27}{3}} = e q o f H_{2}; \frac{4.5}{9} = e q o f H_{2}$
$2 H^{+} + 2 e^{-} \to H_{2}$
eq. of $H_{2}$ = Number of moles $\times n$ factor $0.5 = n_{H_{2}} \times 2$
$V_{H_{2}} = \frac{0.5}{2} \times 22.4; V_{H_{2}} = 5.6 L$

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4.5 g

of aluminum (at. mass 27amu) is deposited at cathode from

A l^{3 +}

solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from

H +

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of aluminium (At. mass

= 27 a t m

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Q. 4.5 g of aluminium (at. mass 27 amu) is deposited at cathode from

A l^{3 +}

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:

$4.5 g$ of Aluminium is deposited at cathode from $A l^{3 +}$ solutions 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:

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4.5g of aluminum (at. mass 27amu) 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

The correct option is C 5.6 LEq of Al=eq of H2 4.5273=eq of H2; 4.59=eq of H2 2H++2e−→H2 eq. of H2= Number of moles × n factor 0.5=nH2×2 VH2=0.52×22.4; VH2=5.6 L

$4.5 g$ of aluminum (at. mass 27amu) is deposited at cathode from $A l^{3 +}$ 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