Question

find no. of electrons in

(a) 3.6 grams of ammonium ion

(b)3.2 grams of methane

Answer 1

This is an easy one, but requires thinking :P 
a) No. of electrons in 3.6 gm of ammonium ion. 

Ans. Ammonium Ion is (NH4)$N H_{4} ^{+}$
This means that Ammonium Ion is formed by losing one electron. 
No. of Protons is equal to Atomic number of N + 4 times atomic number of Hydrogen.
=> No. of protons in NH4 is 7 + 4 = 11
SInce NH4 is positively charged, there should be 10 electrons [so that it is plus 1 positively charged]
=> 1 atom of NH4 contains 10 electrons.
SInce 1 mole contains NA [Avogadro's number] number of ions of NH4,
1 mole of NH4 will contain NA*10 electrons. 
3.6 gm of Ammonium ion is 3.6/18 moles of NH4 = 1/5 moles = 0.2 moles of NH4. 
1 mole of NH4 contains NA*10 e-.
=> 0.2 moles of NH4 contains NA*10*0.2 e-
= 2NA electrons = $12.044 * 10^{23} electrons of N H_{4} ^{+}$ 

b) I hope you are a true science student, and can find the same for 3.2 gm of Methane, that will be $0.2*[6.022* 10^{23}] * 10$ 
The above is the no. of electrons of 3.2 gm of CH4.
I think we have got same answers, isn't it ? :-) 

Answer 2

(a)3.6 gms of NH4+ means 3.6/18 moles=0.2 moles
no. of electrons in 3.6 gms of NH4+=0.2*6.023*10^23*10
  (There are 10 electrons and 11 protons in NH4+).              [no. of moles=given wt./G.M.W.]
                                                         =12.046*10^23 e-
(b)3.2 gms of CH4 means 3.2/16 moles=0.2 moles
no. of electrons in 3.2 gms CH4=0.2*6.023*10^23*10
  (there are 10 electrons and 10 protons in CH4)
                                                   =12.046*10^23 e-