Answer:
30.8% B¹⁰
Explanation:
Given data:
Molar mass of boron = 10.76 g/mol
Molar mass of B¹⁰ = 10.013 g/mol
Molar mass of B¹¹ = 11.093 g/mol
Percentage abundance of B¹⁰ = ?
Solution:
B¹⁰ = x
B¹¹ = 1-x
The equation is,
B¹⁰x + B¹¹(1-x) = 10.76
10.013 x + 11.093 (1-x) = 10.76
10.013 x + 11.093 - 11.093 x = 10.76
10.013 x - 11.093 x = 10.76 - 11.093
-1.08 x = -0.333
x = -0.333/-1.08
x = 0.308
0.308 ×100 = 30.8%
30.8% B¹⁰
100-30.8 = 69.2 %
69.2 % B¹¹
