Mole concept ## Mole concept –

A mole means a collection of 6.023 x 1023 (Avogadro’s number) entities. These entities may be atoms, molecules, ions, electrons, protons as there are in 12 grams of Carbon-12. source : Tes

1 gram atom = N – atoms = 6.023 x 1023 atoms = gram atomic weight

gram atomic weight = weight of  N atoms in grams

1 gram molecule = N- molecules = 6.023 x 1023 molecules = gram molecular wt.

gram molecular wt. = wt. of N molecules in gram

number of moles of molecules ‘n’ = wt. (gram)/ molecular wt.= w/ m

number of moles of atoms ‘n’ = wt. (gram)/ atomic wt.

no. of moles ‘n’ = no. of particles / ( 6.023 x 1023) source : pieehsp.com

Note – Particles may be atoms , molecules, electrons , protons etc. source : SimplyLearnt.com source : Sriharsha ghanta

At NTP,

no. of moles = volume (litre)/22.4 l

#### Solution –

6.023 x 1023 atoms = 1 gram atom

1  atom = 1/ 6.023 x 1023

1 atom = 1.66 x 10-24 gram atom

#### Solution –

wt. of N atoms in grams = gram atomic weight

wt. of 6.023 x 1023 atoms of hydrogen = 1 gram

wt. of 1 atom of hydrogen =1/ 6.023 x 1023 = 1.66 x 10-24

hence mass of 1 atom of hydrogen is 1.66 x 10-24 gm.

OR

n = no. of atoms / 6.023 x 1023

n= 1 / 6.023 x 1023

atomic wt. of Hydrogen = 1

n =  wt./atomic wt.

1 / 6.023 x 1023  = wt./1

wt. = 1.66 x 10 -24 gm. Ans.

#### Solution-

Mass of 1 gm. molecule of NaCl = gram molecular wt. of NaCl

mass of 1 mole of NaCl = gram molecular  weight  of NaCl

= 23 + 35 .5 = 58.5 Ans.

OR

n = wt. / molecular wt.

n =1 , mol.wt. =58.5

1 = wt./ 58.5

wt. =58.5 Ans.

#### Solution-

w= 10 grams

m of C12H22O11 = 12 x 12 + 22 x 1 + 11 x 16 =342

no. of moles ‘n ‘ = w(gm.)/ molecular wt.

n= 10/342= 0.0292 Ans.

#### Solution-

i) No. of gm. atoms or no. of moles ‘n’= wt. / at.wt.= 4.6/ 23 =0.2

n = no. of atoms/ 6.023 x 1023

no. of atoms = n x 6.023 x 1023

no. of atoms = 0.2 x 6.023 x 1023

no. of atoms = 1.2046 x 1023 Ans.

ii) No. of gm. atoms or no. of moles ‘n’= wt. / at.wt.= 60/ 12 =5

n = no. of atoms / 6.023 x 1023

no. of atoms = n x 6.023 x 1023

no. of atoms = 5 x 6.023 x 1023

no. of atoms = 30.115 x 1023 Ans.

iii)    No. of gm. atoms or no. of moles ‘n’= wt. / at.wt.= 72.52/ 207.2 = 0.35

n = no. of atoms/ 6.023 x 1023

no. of atoms = n x 6.023 x 1023

no. of atoms = 0.35 x 6.023 x 1023

no. of atoms = 2.11 x 1023 Ans.

#### Solution –

1 year = 365 x 24 x 60 x 60 sec.

Total rupees to be spent per year= 106 x 3600 x 24 x 365

106 x 3600 x 24 x 365  rupees spent in = 1 year

6.023 x 1023 rupees spent in = ( 6.023 x 1023) / (106 x 3600 x 24 x 365)

= 1.91 x 1010  years Ans.

#### Solution –

no. of gm. atoms or no. of moles ‘n’ = no. of atoms / 6.023 x 1023

n = (2 x 1023)/ (6.023 x 1023)

n =0.33 mole

n = wt. of atom / atomic weight

0.33 = wt. of atom / 32

wt. of atom = 0.33 x 32= 10.56 gm. Ans.

#### Solution –

Mol. wt. of O2= 32, wt. = 64 gm.

moles of O2 ‘n’ = wt. / mol.wt.

‘n’= 64/32 =2 Ans.

n= no. of molecules / 6.023 x 1023

2= no. of molecules / 6.023 x 1023

no. of molecules = 2 x 6.023 x 1023

= 12.046 x 1023 Ans.

Wt. of 6.023 x 1023 molecules=Mol.Wt. = 32

Wt. of one molecule =32/(6.023 x 1023)

Wt. of one molecule = 5.313 x 10-23 Ans.

#### Solution –

Mass of 6.023 x 1023 atoms =at.wt.= 23

Mass of one Na atom = 23/ (6.023 x 1023) = 3.82 x 10-23 gm. Ans.

OR

n = no. of atoms/ 6.023 x 1023

n= 1 / 6.023 x 1023

n =  wt. / At. wt.

at.wt. =23

1 / 6.023 x 1023 = wt./23

wt. = 23 / 6.023 x 1023 = 3.82 x 10-23 gm. Ans.

#### Solution –

No. of moles ‘ n’ = no of molecules / 6.023 x 1023

n =( 6.023 x 1022) / 6.023 x 1023 = 0.1 mole

Mol.wt. of CO2= 12+ 16 x 2=44

n= wt. of molecules / mol.wt.

0.1 =wt. of molecules / 44

wt.of molecule = 44 x 0.1 = 4.4 gm.Ans.

#### Solution –

no. of moles ‘n’ = w/m

w =500 gm. , m = 2 +16 =18

n = 500 /18 =27.78 moles Ans.

n= no. of molecules / 6.023 x 1023

27.78 = no. of molecules / 6.023 x 1023

no. of molecules = 27.78 x 6.023 x 1023

no. of molecules= 1.67 x 1025 Ans.

No. of electrons in water (H2O) = 2 +8 =10

no. of electrons in one molecule of water = 10

no. of electrons in  1.67 x 1025  molecule of water =10 x 1.67 x 1025

= 1.67 x 1025  Ans.

#### Solution –

no. of moles ‘n’ = w/m

w = 168 gm. , m = 24 +12 + 48 = 84

n = 168 /84 = 2 moles Ans.

n= no. of molecules / 6.023 x 1023

2 = no. of molecules / 6.023 x 1023

no. of molecules = 2 x 6.023 x 1023

no. of molecules= 12.046 x 1023 Ans.

No. of atoms in MgCO3 = 1 +1 + 3 = 5

no. of atoms in one molecule of MgCO3  = 5

no. of atoms in  12.046 x 1023  molecule of water = 12.046 x 5 x 1023

= 60.23 x 1023  Ans.

#### Solution –

no. of moles ‘n’ = w/m

w = 1.06 gm. , m = 23 x 2 + 12 + 48 = 106

n = 1.06 /106 = 0.01 mole Ans.

n= no. of molecules / 6.023 x 1023

0.01 = no. of molecules / 6.023 x 1023

no. of molecules = 0.01 x 6.023 x 1023

no. of molecules= 6.023 x 1021 Ans.

No. of ions in  Na2CO3 = 2Na+ +CO3 =3

no. of ions in one molecule of Na2CO3 = 3

no. of ions in  6.023 x 1021  molecule of water =3 x 6.023 x 1021

= 18.069 x 1021  Ans. 