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Asked by choudharylucky155 | 22 Apr, 2023, 05:06: PM
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Atomic weight of iron = 55.845 amu ;
 
Number n of atoms in 2.24 mg iron is
 
begin mathsize 14px style n space equals space fraction numerator 2.24 space asterisk times space 10 to the power of negative 3 end exponent over denominator 55.845 end fraction cross times N end style
 
Where N = 6.02 ×  1023 is avagadro number .
 
begin mathsize 12px style n space equals space fraction numerator 2.24 space cross times space 10 to the power of negative 3 end exponent space cross times space 6.023 space cross times space 10 to the power of 23 over denominator 55.845 end fraction space equals space 2.416 space cross times space 10 to the power of 19 end style
Number of electrons removed , ne = 2.416 × 1019 × 0.02 × 10-2 = 4.832 × 1015 
 
Charge removed from given quantity of iron = -4.832 × 1015 × 1.602 × 10-19 = -7.741 × 10-4 C 
 
Charge on given 2.24 mg iron = + 7.741 × 10-4 C
 
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Answered by Thiyagarajan K | 22 Apr, 2023, 08:41: PM
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