Answer:
Average atomic mass of the vanadium = 50.9415 amu
Isotope (I) of vanadium' s abundance = 99.75 %= 0.9975
Atomic mass of Isotope (I) of vanadium ,m= 50.9440 amu
Isotope (II) of vanadium' s abundance =(100%- 99.75 %) = 0.25 % = 0.0025
Atomic mass of Isotope (II) of vanadium ,m' = ?
Average atomic mass of vanadium =
m × abundance of isotope(I) + m' × abundance of isotope (II)
50.9415 amu =50.9440 amu× 0.9975 + m' × 0.0025
m'= 49.944 amu
Explanation: