Mole concept
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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.
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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)
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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
Question 1- How many gram-atoms are there in one atom ?
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
Question 2 – Calculate the mass of 1 atom of hydrogen?
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.
Question 3 – What is the mass of one mole NaCl?
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.
Question 4- Calculate number of moles in 10 grams of C12H22O11
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.
Question 5- How many atoms & gm. atoms are there in
i) 4.6 gm. Na (ii) 60 gm. Carbon (iii) 72.52 gm. lead
(Given At. wt. of Na , C & Pb are 23 , 12 , 207.2 respectively)
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.
Question 6- How many years would it take to spend Avogadro’s number of rupees at the rate of one million or 10 lakh rupees/ sec?
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.
Question 7 – Find the no. of gm. atoms & wt. of an element having 2 x 1023 atoms. At. wt. of element is 32.
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.
Question 8 – How many moles & molecules of O2are there in 64 gm. O2. What is the mass of one molecule of O2 ?
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.
Question 9 – If atomic mass of Na atom is 23 . Find mass of one Na atom?
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.
Question 10 – What is the mass of 6.023 x 1022 molecules of CO2 ?
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.
Question 11 – How many moles,molecules & electrons in 500 gm.water ?
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.
Question 12 – How many moles,molecules & atoms in 168 gm. MgCO3 ?
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.
Question 13 – How many moles,molecules & ions in 1.06 gm. Na2CO3 ?
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.
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