Question

Suppose the isotopic ratio of the two boron isotopes 10B (10.013 amu) and 11B (11.009 amu) in a sample has been altered from the ratio found in nature and now contains 33.36% 10B in the sample. Determine the atomic weight of this sample of this new boron element.

Answer 1

The correct answer is 10.676 amu.

Step-by-step explanation:

Based on the given information, the concentration of 10B left in the sample is 33.36%. Therefore, the percentage of 11B present will be,

11B = 100% - 33.36% = 66.64%

Now the atomic weight of the new boron element can be determined by adding the atomic masses of both the isotopes multiplied by its percentage.

Therefore,

= (10.013 amu * 33.36%) + (11.009 amu * 66.64%) / 100

= 10.676 amu