A piece of aluminum has a mass of 27g and is initially electrically neutral. It then acquires a...

Question:

A piece of aluminum has a mass of 27g and is initially electrically neutral. It then acquires a charge of q=0.5uC by removing electrons from it. The molar weight of aluminum is 27g and its atomic number is 13.

a) How many electrons were removed from the aluminum piece in the charging process?

b) What percentage of the total number of electrons initially present in the aluminum piece does the number in part (a) represent?

Charged Due to Loss of Electrons.

The Avogadro's number helps us determine the number of elementary charges (generally electrons) in an element from its molar mass when its mass is provided. The number of charges in an element is quantized and is integral multiples of the elementary charge.

Part(a.)

Given data,

{eq}\textrm{Mass of Aluminum,}\; m = 27\;\mathrm{g}\\ \textrm{Atomic Number,}\; Z = 13\\ \textrm{Molar mass of Aluminum,}\;M = 26.98\;\mathrm{g/mol} {/eq}

Therefore, the total number of moles in the given mass of Aluminum is, {eq}\dfrac{27}{26.98} = 1.00074\;\mathrm{mol} {/eq}

By Avogadro's law, we can determine the number of electrons present in the 1.00074 moles of Aluminum.

Mathematically,

{eq}1.00074\times 6.022\times 10^{23} = 6.026\times 10^{23}\;\mathrm{electrons}.......(1) {/eq}

Using the law of quantization of charge we can determine the number of electrons lost to gain the given positive charge.

Mathematically,

{eq}q = ne\\ \textrm{Here, putting the given values,}\\ 0.5\times 10^{-6} = n (1.6\times 10^{-19})\\ n = 3.125\times 10^{12}\;\mathrm{eletrons}......(2) {/eq}

Part(b.)

Using (1) & (2) we obtain the loss percentage as,

{eq}Loss \% = \dfrac{3.125\times 10^{12}}{6.026\times10^{23}}\; \times 100\\ Loss \% = 0.518\times 10^{-9} {/eq} 