Question

(i) Calculate the total number of electrons present in one 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 = \[1.675\times {{10}^{-27}}kg\]).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of \[N{{H}_{3}}\]at STP.
Will the answer change if the temperature and pressure are changed?

Answer 1

Hint:. Mole is a quantity which measures the amount of substance present in the given sample. Mole can be measured in the terms of atoms, ions, electrons, etc or the form of mass.

Complete step by step answer:
(i) 1 mole of Methane = 1 mole of Carbon + 4 moles of Hydrogen
Therefore, 1 mole of C = 6 mole of electrons
1 mole of H = 1 mole of electrons,
There are 10 moles of electrons.
But, 1 mole is also \[6.022\times {{10}^{23}}\] electrons
Hence, 1 mole of methane = \[10\times 6.022\times {{10}^{23}}\] electrons
= \[6.022\times {{10}^{24}}\]electrons

(ii) (a) Given weight of \[^{14}C\] = 7 mg
Gram atomic mass = 14g/mol
No. of moles = Given wt./ Atomic weight
= 7mg/14g/mol
= \[0.5\times {{10}^{-3}}\] mol
No. of neutrons in one Carbon atom = 8
No. of moles of neutrons = no. of moles of Carbon × 8
= \[0.5\times {{10}^{-3}}\times 8\]
= \[4\times {{10}^{-3}}\]

(b) Total number of neutrons = no. of moles of neutron × Avogadro’s number
= \[4\times {{10}^{-3}}\times 6.022\times {{10}^{23}}\]
= \[24.022\times {{10}^{21}}\]

(iii) Given wt. of \[N{{H}_{3}}\]= 34 mg
Molecular mass of \[N{{H}_{3}}\]= 17g/mol
No. of moles = 34mg/ 14g/mol
=\[2\times {{10}^{-3}}\] mol
1 mole of \[N{{H}_{3}}\] = 10 moles of protons
Hence, no. of moles of \[N{{H}_{3}}\]in 34 mg = \[10\times 2\times {{10}^{-3}}\]
 = \[2\times {{10}^{-2}}\]

(a) To find the no. of protons = no.of mole of protons × Avogadro’s number
= \[2\times {{10}^{-2}}\times 6.022\times {{10}^{23}}\]
= \[1.2044\times {{10}^{22}}\]protons

(b) Mass of protons= no. of protons × mass of 1 proton
= \[1.2044\times {{10}^{22}}\times 1.6726\times {{10}^{-27}}\]kg
= \[2.01447\times {{10}^{-5}}\]kg
No, the answer won’t change if the temperature and pressure are changed.

Note: This answer can be done via short method but it is also important to know the basics of the question. Also, there is gram-atomic mass used in this question instead of the atomic mass because the unit of gram-atomic mass is gm/mole which makes 1 mole. Therefore, keep in mind the units of atomic and gram atomic mass.