Question

answers Batao friends first​

Answer 1

Answer:

Given:

Average atomic mass = 16.2 u

The isotopes have been given.

To find:

% of the isotopes present in nature.

Calculation:

Let % of X(16) be "y"

Then , % of X(18) be "100 - y"

$avg.mass \: = \frac{16y \: + 18(100 - y)}{100}$

$= > 16.2 = \frac{16y \: + 18(100 - y)}{100}$

$= > 1620 = 16y \: + \: 18(100 - y)$

$= > 1620 = 16y \: + \: 1800 - 18y$

$= > 2y = 180$

$= > y = 90$

Hence,

the X(16) isotope is present 90% ;

the X(18) isotope is present 10 %

Answer 2

Solution

Given

The average atomic mass of the element X is 16.2 u

To find

Percentage of $\sf{X^{16}} and\ \sf{{X}^{18}}$

Let the percentage composition of one of the isotope having mass 16u be y

• Composition of isotope with 18u will be (100 - y)

Mathematically,

$\large{ \sf{16.2 =  \dfrac{16y + 18(100 - y)}{100} }} \\  \\  \large{ \leadsto \:  \sf{1620 = 16y - 18y + 1800}} \\  \\  \large{  \leadsto \: \sf{ - 180 =  - 2y}} \\  \\  \large{ \leadsto \:  \sf{y =  \frac{180}{2} }} \\  \\  \huge{ \leadsto \boxed{ \boxed{\sf{ \blue{y = 90}}}}}$

Thus,the percentage composition of 16 u isotope is 90 percent and 18 u isotope is 10