Question

Amount of zinc deposited when $3.01\times {10}^{20}$ electrons pass through $ZnS{O}_4$ solution is:

• A. $16.35g$
• B. $16.375\times {10}^{-3}g$
• C. $32.7\times {10}^{-3}g$
• D. $32.7g$

Answer 1

The correct option is B $16.375\times {10}^{-3}g$

$Z{n}^{+2}+2e^{-}\to Zn$

$6.02\times {10}^{23}$ electrons $\to$ 1 mol electrons $\to 0.5$ moles $Zn$.

$3.01\times {10}^{20}$ electrons $\to 5\times {10}^{-4}$ mole of electrons $=2.5\times {10}^{-4}$ mol of Zn

Molar mass of $Zn=65.5g/mol$

Mass of Zn deposited $=moles\times molarmass=2.5\times {10}^{-4}\times 65.5=16.375\times {10}^{-3}g$

Option B is correct.