Chemistry 2001 JAMB Past Questions - Question 10: An element X with relative atomic mass 16.2 contains two isotopes 168X with relative abundance of 90% and m8X with relative abundance of 10%. The value of m is

Correct Answer

B

Explanation

90/100 X 16 + 10/100 X m(14.4) + (0.1 x m) = 16.20.1 x m = (16.2 - 14.4) = 0.1.x m = 1.8m=1.8/0.1 = 18
The relative atomic mass (Ar) of an element is the weighted average of the masses of its isotopes, taking into account their relative abundances. The formula for calculating the relative atomic mass is:

\[ Ar = (A_1 \times X_1) + (A_2 \times X_2) + \ldots \]

where:
- \( A_1, A_2, \ldots \) are the atomic masses of the isotopes,
- \( X_1, X_2, \ldots \) are their relative abundances (expressed as decimals).

In this case, element X has two isotopes: \( ^{16}_{8}X \) and \( ^{m}_{8}X \), with relative abundances of 90% and 10%, respectively.

Let's denote the atomic mass of \( ^{16}_{8}X \) as 16 and the atomic mass of \( ^{m}_{8}X \) as \( m \). The relative atomic mass (Ar) is given as 16.2. Now we can set up the equation:

\[ 16.2 = (16 \times 0.9) + (m \times 0.1) \]

Solving for \( m \):

\[ 16.2 = 14.4 + 0.1m \]

Subtracting 14.4 from both sides:

\[ 0.1m = 1.8 \]

Dividing by 0.1:

\[ m = 18 \]

Therefore, the value of \( m \) is 18.