Question

a) The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes 168X and 188X in the sample?

b)Give reasons:
(i) Isobars of an element show different chemical properties
(ii) Al+3 has no electron in M shell. (atomic no. of Al =13)

Answer 1

Answer:

Suppose the percentage of  

  16

X  in given sample = x8

:.  The percentage of  

18

X  in given sample = 100-x8

  

 As given average atomic mass of a sample of an element X = 16.2 u

=> (16  × x /100) + [18  × (100-x) /100] = 16.2

 => (16x + 1800-18x) / 100 = 16.2

=>  (16x + 1800-18x) = 16.2 × 100

=>  -2x + 1800 = 1620

=> -2x = 1620-1800

= -2x = -180

=> x = 90

:.  The  percentage of  

16

X  in given sample = x = 90 8

:.  The  percentage of  

16

X  in given sample = 100-x =100-90 = 108

 

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