The answer is 10.81 u
Solution :
Relative abundance of 10B isotope = 19.60%
Relative abundance of 11B isotope = 80.40%
\(Average\space atomic \space mass= {(Mass \space of\space isotope \space 1) \times
(
Relative\space abundance \space of \space ^{10}B \space isotope)
+ (Mass \space of\space isotope\space 2) \times (Relative \space abundance
of\space ^{11}B \space isotope \over 100}\)
=\((10 \times 19.6 ) + (11 \times 80.4 ) \over100\)
\(={196+ 884.4\over 100}={1080.4\over 100}\)
=10.804 u