Physical chemistry–A-1

Question: – calculate the total n of atoms in 78gm of k
Solution: – 6.022×1023 atom =39gm of k
78gm of k = 6.022×1023 atom k=12.044×1023 atom
Question: – Calculate the total no of decimoles in 46gm of Na
Solution: – 1 mole of Na = 23gm
46gm = 2mole
1mole = 10decimole
2mole =20decimoele
Use of mole concept of compound·
Particle = molecule
Formula wt. =mol wt.
Name of mole =gm mole or, gm molecular
One gm mole CaCo3 = one mole CaCo3
6.022×1023 /H1H2 molecular = 100gm
One gm –atom of hydrogen = 1gm
one gm –mole of hydrogen = 2gm
one mole Na2Co3 = one gm mole of Na2Co3 =6.022×1023 molecular =106gm
total no of e- ion one molecular of NaCo3 =6atom
one molecular of NaCo3 =2atomof Na = 1atom of Ca = 3atom of o
no of e- in one molecular of NaCo3 = 11×2+6+8×3 =52e-
Question: – calculate the total no of atom in 49gm f H2So4
Solution: – one mole H2So4 = 6.022x1023molecular =98gm
49gm = 6.022×1023/2 molecular
= 3.022x1023molecular
Total no atom of in H2So4 = 7atom
Question: -calculate the total no of e- in 50gm caco3
Solution: – one mole caco3 = 6.022x1023molecular
50gm = 6.022x1023x56/100= 3.011x1023moecular
Total no e- in ca = 20

Page no.- 04
C = 6
O = 24
Total no of e-in = 3.022x1023x50e-
Use of me concept for radicle·
Particle = ion
Formula wt. = ionic wt.
Moe = gm – ion
One mole Na+=one gm –ion Na+ = 6.022x1023ion = 23gm
Na Na+
11e- 10e-
11p 11p
12N 12p
A = 11p+12p A = 11p+12n
No of electron in one ion of Na+ =10e-
Question – calculate the total no of e-in 48gm of So4.

Solution – one mole of So4 = 32+16×4
= 96gm =6.022×1023
=48gm = 3.022x1023ion
No of e- one ion of So4 =16+8×4+2 = 50e-
Mole in terms of volume (only for gaseous substance) ·
Molar volume: – volume of one mole nature gas at any temperature and
pressure is called molar volume.
Case1: –
Molar volume of S.T.P or N.T.P
T =00C or 273k p = 1atom
Molar volume at S.T.P = 22.4 lit
=22400cc(or) ml
One gm-mole oxygen at S.T.P = 22.4 lit



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