Ar of Zirconium, Ar of Iron, Ar of Strentium, PROBLEM A Level, O level Chemistry Notes
SOLUTION:
Ar of Zirconium,
(90X51) + (91X11.2) + (92X17.1) + (94.17.4) +
(96X2.8)
100
= 9131.8 / 100
= 91.318
Ar of Iron
=
(54x10) +
(56x90)
100
= (540) + (5040)
100
= 5580 / 100
= 55.8
Ar of
Strentium,
= (86x20) + (88x80)
100
= (1720) + (7040)
100
= 8760 / 100
= 87.6
PROBLEM:
Selenium, Se is in group iv of the
periodic table and occurs in nature as a mixture of six isotopes, having the
relative atomic mass & relative abundance as given below:
|
Mass
Number |
74 |
76 |
77 |
78 |
80 |
82 |
|
%
Abundance |
0.9 |
9.0 |
7.6 |
23.5 |
49.8 |
9.2 |
SOLUTION
Ar of
Selenium, Se
(74x0.9) +
(76x9.0) + (77x7.6) + (78x23.5) + (80x49.8) + (82x9.2)
100
(66.6) + (684) + (585.2) + (1833) + (3984) + (754.4)
100
= 7907.2 /
100
= 79.072
PROBLEM:
The relative atomic mass of potassium is
39.1. The two main isotopes of potassium are 39K and 41K.
Which isotope is more abundant? What is the proportion of each isotopes in the
sample of potassium?
SOLUTION:
Ar of
potassium = 39.1
% abundance
of 39K = P%
% abundance
of 4.1K = (100 – P) %
39.1 = (39xP) + 41(100-P) / 100
3910 = 39P + 4100 – 41P
41P – 39P = 4100 – 3910
2P = 190
P = 190 / 2
P = 95
% abundance
of 39K = 95%
% abundance
of 41K = 5%
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