An elements X with relative atomic mass 16 2 co... - JAMB Chemistry 2001 Question
An elements X with relative atomic mass 16.2 contains two isotopes 168X with relative abundance of 90% and m8X with relative abundance of 10%. The value of m is
A
16
B
18
C
12
D
14
correct option: b
168X, m8X
16 * (90/100) + m * (10/100) = 16.2
1440 + 10m = 1620
10m = 180
m = 18
16 * (90/100) + m * (10/100) = 16.2
1440 + 10m = 1620
10m = 180
m = 18
Please share this, thanks:
Add your answer
No responses