Question

Calculate the charges of an iron particle of mass 224 mg if 0.01% of the electrons are removed from it.

Answer 1

mass number of Fe = 56 g

moles of Fe = 22.4/56 =0.4

no of electrons in 0.4 moles= 0.4*6.023*10 to power 23 =2.4092*10 to power 23

when .01% electrons are removed then left no of electrons =2.4089*10^23

charge on iron 2.4089*10^23*1.6*10^-19= 38543.34 C

Answer 2

Answer:

The positive charge will be $0.096\times10^{4}\ C$

Explanation:

Given that,

Mass of iron particle = 224 mg

Atomic mass of iron = 56

One mole of iron contains 56 g of iron.

Number of moles in 224 g iron is

$n= \dfrac{224}{56}$

$n= 4$

Total numbers of electrons or protons in 4 moles of iron

Total number =$n\times N$

Here, N= Avogadro number

Total number = $4\times26\times 6.032\times10^{23}$

a = 0.01 % of total number

$a=\dfrac{0.01}{100}\times4\times26\times6.032\times10^{23}$

$a= 0.063\times10^{23}$

$The\ negative\ charge = a\times\ charge\ electron$

The negative charge = $0.063\times10^{23}\times1.6\times10^{-19}$

The negative charge = $0.096\times10^{4}\ C$

The positive charge = $0.096\times10^{4}\ C$

Hence, The positive charge will be $0.096\times10^{4}\ C$