(i) Calculate the total number of electrons present in 1 mole of methane .
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of `.^(14) C`. (Assume that mass of a neutron ` = 1 . 6 75 xx 10 ^(-27) g)`
(iii) Find (a) the total number of protons and (b) the total mass fo protons in ` 32 mg` of ` NH_3` at ` STP`. ( mass of proton ` = 1 . 6 72 xx 10^(-27) g)`
Will the answer change if the temperature and pressure are changed ?

Explanation

(i) One mole of methane `(CH_(4))` has molecules `=6.022 xx 10^(23)`
No. of electrons present in one molecule of `CH_(4)=6+4=10`
No. of electrons presents in `6.022 xx 10^(23)` molecules of `CH_(4)= 6.022 xx 10^(23) xx 10`
`=6.022 xx 10^(24)` electrons
(ii) Step I. Calculation of total number of carbon atoms
Gram atomic mass of carbon (C-14)`=14 g = 14 xx 10^(3) mg`
`14 xx 10^(3) mg` of carbon (C-14) have atoms `= 6.022 xx 10^(23)`
7 mg of carbon (C-14) have atoms `=(6.022 xx 10^(23))/((14 xx 10^(3)mg)) xx (7 mg)= 3.011 xx 10^(20)` atoms.
Step II Calculation of total number and total mass of neutrons
No. of neutrons present in one atom (C-14) of carbon = 14 -6 =8
No. of neutrons present in `3.011 xx 10^(20) ` atoms (C-14) of carbon = `3.011 xx 10^(20) xx 8`
`=2.408 xx 10^(21)` neutrons
Mass of one neutrons `= 1.675 xx 10^(-27)` kg
Mass of `2.408 xx 10^(21) ` neutrons `= (1.675 xx 10^(-27)kg) xx 2.408 xx 10^(21)`
`=4.033 xx 10^(-6) `kg.
(iii) Step I. Calculation of total number of `NH_(3)` molecules
Gram molecular mass of ammonia `(NH_(3))= 17 g =17 xx 10^(3)`mg
` 17xx10^(3)` mg of `NH_(3)` have molecules `= 6.022 xx 10^(23)`
34 mg of `NH_(3)` have molecules `=(6.022 xx 10^(23))/((17xx10^(3)mg))xx(34 mg)`
`=1.2044 xx 10^(20) ` molecules.
Step II. Calculation of total number and mass of protons
No. of protons present in one molecule of `NH_(3)=7+3=10`
No. of protons present in `12.044 xx 10^(20) ` molecules of `NH_(3)=12.044 xx 10^(20) xx 10`
`=1.2044 xx 10^(22)` protons
Mass of one proton `=1.67 xx 10^(-27)kg`
Mass of `1.2044 xx 10^(22)` Protons `=(1.67 xx 10^(-27) kg) xx 1.2044 xx 10^(22)`
`=2.01 xx 10^(-5) kg`.
No, the answer will not change upon changing the temperature and pressure because only the number of protons and mass of protons are involved.