JAMB - Chemistry (2015 - No. 57)
Chlorine consisting of two isotope of mass number 35 and 37 in the ratio 3:1 has an atomic mass of 35.5. Calculate the relative abundance of the isotope of mass number 37
20
25
60
75
Explication
Let the relative abundance of the energy with mass 37 be
Total relative abundance of the isotope of mass 359100-k)
Total mass of isotope of mass 37-37x, while that of 35-359(100x)
-37x + 3500 − 35 − 35.5
-37x + 3500 − 35n − 35.5
Mean mass of chlorine atom − (2n + 3500)/ ÷ 100
= 35.5
2x + 3500 − 3550yx = 50/2
Total relative abundance of the isotope of mass 359100-k)
Total mass of isotope of mass 37-37x, while that of 35-359(100x)
-37x + 3500 − 35 − 35.5
-37x + 3500 − 35n − 35.5
Mean mass of chlorine atom − (2n + 3500)/ ÷ 100
= 35.5
2x + 3500 − 3550yx = 50/2