Question

Atomic weight of boron is 10.81 and it has two isotopes
₅B¹⁰ and ₅B¹¹. Then ratio of ₅B¹⁰ : ₅B¹¹ in nature would
be
(a) 19 : 81 (b) 10 : 11
(c) 15 : 16 (d) 81 : 19

Answer 1

$\maltese$ Answer $\maltese$

$\checkmark$ Given:

⇒The Atomic Weight of Boron is 10.81

⇒ It has Two Isotopes

• ₅ B ¹⁰
• ₅ B ¹¹

$\checkmark$ To Find:

⇒ Ratio of ₅ B ¹⁰ : ₅ B ¹¹

$\checkmark$ Solution:

We know that,

Average Atomic Mass

It is defined as the sum of the masses of its isotopes multiplied to its natural occurrence

Given that,

The Atomic Weight of Boron = 10.81

So,

Let as assume,

Percentage (%) of B ¹⁰ to be $\bold{\rm k}$

Hence,

The Average Atomic Weight is

$\orange{\implies \rm \dfrac{10(k)+11(100-k)}{100}=10.81}$

$\red{\implies \rm 10(k)+11(100-k)=1081}$

$\green{\implies \rm 10k+1100-11k=1081}$

$\blue{\implies \rm 11k-10k=1100-1081}$

$\pink{\implies \rm k=19}$

Therefore,

We have taken,

₅ B ¹⁰ = k

So,

₅ B ¹⁰ = 19

And, We have taken,

₅ B ¹¹ = 100 - k

So,

₅ B ¹¹ = 100 - 19

₅ B ¹¹ = 81

According to the Question,

We are asked to find

The ratio of ₅ B ¹⁰ : ₅ B ¹¹ in nature

Hence,

$\red{\implies \rm \dfrac{ _{5}B^{10}}{ _{5}B^{11}}=\dfrac{19}{81} }$

Therefore,

₅ B ¹⁰ : ₅ B ¹¹ = 19 : 81

Hence,

$\green{\checkmark \ \rm Option \ (a) \ is \ correct }$