005
10.0 points
the molar mass of boron atoms in a natural
sample is 10.76 g/mol. the sample is known
to consist of "b (molar mass 10.013 g/mol)
and b (molar mass 11.093 g/mol). what is
the percentage abundance of 10b?

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

1.solid

2.carbon dioxide

3.helium