The number of moles of electrons required to deposit 36 g of Al from an aqueous solution of Al ( NO 3)3 is (At. wt. of Al =27 )

Question

The number of moles of electrons required to deposit 36 g of Al from an aqueous solution of Al ( NO 3)3 is (At. wt. of Al =27 ) (A) 4 (B) 2 (C) 3 (D) 1

Tardigrade Admin · Accepted Answer

Al 3++ underset text 3 molO3 e- -> underset 1 mol =27 gAl ∵ 27 g of Al is deposited by 3 moles of electrons ∴ 36 g Al will be deposited by electrons =(3/27) × 36=4 mol

Q. The number of moles of electrons required to deposit $36 g$ of $A l$ from an aqueous solution of $A l (N O_{3})_{3}$ is (At. wt. of $A l = 27$ )

4

2

3

1

$A l^{3 +} + 3 mol O 3 e^{-} 1 m o l = 27 g A l$

$∵ 27 g$ of $A l$ is deposited by $3$ moles of electrons

$∴ 36 g A l$ will be deposited by electrons

$= \frac{3}{27} \times 36 = 4 m o l$

Q. The number of moles of electrons required to deposit 36g of Al from an aqueous solution of Al(NO3​)3​ is (At. wt. of Al=27 )

4

2

3

1

Al3++ 3 molO3e−​​1mol=27gAl​ ∵27g of Al is deposited by 3 moles of electrons ∴36gAl will be deposited by electrons =273​×36=4mol

Q. The number of moles of electrons required to deposit $36 g$ of $A l$ from an aqueous solution of $A l (N O_{3})_{3}$ is (At. wt. of $A l = 27$ )

$A l^{3 +} + 3 mol O 3 e^{-} 1 m o l = 27 g A l$

$∵ 27 g$ of $A l$ is deposited by $3$ moles of electrons

$∴ 36 g A l$ will be deposited by electrons

$= \frac{3}{27} \times 36 = 4 m o l$