Question

Eq. 2 2kL Trial # | Charge mass "half R balioon Number excess electrons ク㓁 3,て6x10 2 maximum |.po子 1o.bib

Answer 1

Slight No. of moles = 7/12 = 0.5833 No. of atom = N_A middot m N = 6.023 times 10^23 times 0.5833 N = 3.5134 times 10^23 Now extra electron = 3.26 times 10^N So Portion which have extra electron is = 3.26 times 10^11/3.5134 times 10^23 = 9.278 times 10^-13 Past Portion which have extra electron = 9.23 times 10^4/3.5134 times 10^23 = 2.62 times 10^-12 th Part