What are the percentage abundance of these 2 isotopes?

Question

the molar mass of Boron atoms in a natural sample is 10.81 g/mol. the sample is known to consist of 10B (molar mass is 10.013) and 11B(molar mass is 11.093) .

Answer 1

the reported molar mass is the sum of the weighted averages of the isotopes.  ie.. the sum of each isotope's mass x abundance...

let y = fraction of 10B

1 - y  = fraction of 11B

10.013 y + 11.093 (1 - y) = 10.81

10.013 y + 11.093 - 11.093 y = 10.81

-1.080 y =  - 0.283

y = 0.262 = 26.2%

1-y = 1 - 0.262= 0.738 - 73.8%

answer is

10B is 26.2% , 11B is 73.8%

Answer 2

There are two stables isotopes of boron which comprise 100% of the boron found in the Earth's crust.

B-10 (molar mass is 10.013) --> 19.9%

and

B-11 (molar mass is 11.093) --> 80.1%

I have found the following periodic table to be a very handy resource for isotopic data as well as general physical properties.

http://www.dayah.com/periodic/

Answer 3

10.013x + 11.093(1-x) = 10.81

10.013x + 11.093 - 11.093x =10.81

1.08x = 0.283

x=0.26 (10.013) =26%          1-x=0.74(11.093)=74%

10.013*0.26  +  11.093*0.74 = 10.812

Bye

Answer 4

fraction of B11 =x

10.81= 11x +(1-x)10

10.81=11x+10-10x

.81=x

0.81 B11 and 0.19 B10