The number of electrons required to reduce 4.5 × 10-5 g of Al is

Question

The number of electrons required to reduce 4.5 × 10-5 g of Al is (A)  1.03×1018  (B)  3.01×1018  (C)  4.95×1026  (D)  7.31×1020

Tardigrade Admin · Accepted Answer

underset27Al3++3e- → underset27 gAl 27 g of Al is reduced by =3 × 6.023 × 1023 e- s 4.5 × 10-5 g of Al will be reduced by =(3 × 6.023 × 1023 × 4.5 × 10-5/27) =3.01 × 1018 electrons

Q. The number of electrons required to reduce $4.5 \times 1 0^{- 5}$ g of $A l$ is

$1.03 \times 1 0^{18}$

$3.01 \times 1 0^{18}$

$4.95 \times 1 0^{26}$

$7.31 \times 1 0^{20}$

$27 A l^{3 +} + 3 e^{-} \to 27 g A l$

$27 g$ of $A l$ is reduced by $= 3 \times 6.023 \times 1 0^{23} e^{-} s$

$4.5 \times 1 0^{- 5} g$ of $A l$ will be reduced by

$= \frac{3 \times 6.023 \times 1 0 ^{23} \times 4.5 \times 1 0 ^{- 5}}{27}$

$= 3.01 \times 1 0^{18}$ electrons

Q. The number of electrons required to reduce 4.5×10−5 g of Al is

1.03×1018

3.01×1018

4.95×1026

7.31×1020