A) Given,
Mass of 63Li lithium isotope,
m1=6.01515 u
Mass of 73Li lithium isotope,
m2=7.01600 u
Abundance of 63Li,n1=7.5%
Abundance of 73Li,n2=92.5%
The atomic mass of lithium atom is:
m=m1n1+m2n2n1+n2
Putting values, we get
m=6.01515×7.5+7.01600×92.57.5+92.5
=6.940934 u
Final Answer : 6.941 u
B) Given,
Mass of 105B Boron isotope,
m1=10.01294 u
Mass of 115B Boron isotope,
m2=11.00931 u
The atomic mass of Boron is 10.811 u
Let us assume abundance of 105B
n1=x%
Abundance of 115B,n2=(100−x)%
The atomic mass of Boron atom is:
m=m1n1+m2n2n1+n2
Putting values, we get
10.811=10.01294×x+11.00931×(100−x)x+100−x
1081.11=10.01294x+1100.931−11.00931x
x=19.8210.99637=19.89%
Abundance of 105B,n1=19.89%
Abundance of 115B,n2=(100−x)%
n2=(100−19.89)%
n2=80.11%
Final Answer : 19.9%,80.1%