Question

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?​

Answer 1

30.8% B¹⁰

Step-by-step 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