Question

Naturally occurring Boron consists of two isotopes of atomic masses
10.0 and 11.0.The atomic mass of natural Boron is 10.8.Calculate the
percentage of each isotope in natural Boron.
Pls Explain with step by step

Answer 1

Answer:-

→ Percentage of B₁ isotope = 20%

→ Percentage of B₂ isotope = 80%

Explanation:-

Let the two isotopes of Boron be B₁ and B₂ respectively.

• The mass of B₁ isotope is 10.0 u. Let it's percentage in the sample be x%.

• The mass of B₂ isotope is 11.0 u. Hence, it's percentage in the sample will be (100 - x)%

________________________________

Average atomic mass of 'B' :-

= 10 × x/100 + 11 × (100-x)/100

But as the average atomic mass of Boron is given as 10.8 u, thus :-

⇒ 10 × x/100 + 11 × (100-x)/10 = 10.8

⇒ 10x/100 + 1100-11x/100 = 10.8

⇒ 1100 - x/100 = 10.8

⇒ 1100 - x = 100(10.8)

⇒ 1100 - x = 1080

⇒ -x = 1100 - 1080

⇒ - x = -20

⇒ x = 20

Thus, percentage of B₁ isotope is 20% and percentage of B₂ isotope is (100 - 20) = 80%.

Answer 2

${\bf{Given\::}}$

• Boron consists of two isotopes of atomic masses 10.0 and 11.0.

• The atomic mass of natural Boron is 10.8.

$\\ {\bf{To\: Find\::}}$

• The percentage of each isotope in natural Boron.

$\\ {\bf{Solution\::}}$

Let,

• The percentage of isotope of B with atomic mass 10.0 is x.

Then,

• The percentage of isotopes of B with atomic mass 11.0 be ′100−x′ [ $\because$ in calculating the percentage, the sum is always 100%.]

As we know that,

➣ Average atomic mass can be calculated by,

$\orange\bigstar\:{\Large\mid}\:\bf\purple{Avg.\:At.\:mass\:=\:\dfrac{\left(mass~of~1st~isotope\times{it's~\%}\right)\:+\:\left(mass~of~2nd~isotope\times{it's~\%}\right)}{100}\:}\:{\Large\mid}\:\green\bigstar$

$\dashrightarrow\:\tt{10.8\:=\:\dfrac{10.0\times{(x)}\:+\:11.0\times{(100~-~x)}}{100}\:}$

$\dashrightarrow\:\tt{10x\:+\:1100~-~11x\:=\:10.8\times{100}\:}$

$\dashrightarrow\:\tt{10x\:+\:1100~-~11x\:=\:1080\:}$

$\dashrightarrow\:\tt{10x\:-\:11x\:=\:1080\:-\:1100\:}$

$\dashrightarrow\:\bf{x\:=\:20\:}$

Hence,

➠ % of B with atomic mass 10.0 is 20%.

And,

➠ % of B with atomic mass 11.0 is (100 - 20)% = 80%

∴ % of B with atomic mass 10.0 is $\bf\pink{20\%}$ & % of B with atomic mass 11.0 is $\bf\pink{80\%}$.