Question

1) Assume element X has 2 isotopes: X-125 and X-126. For every 100 atoms of X, 30 of them have a mass of 125.0 u and 70 have a mass of 126.0 u. What is the average atomic mass of X?
A)125.3 u B)125.7 u C)126.3 u D)126.7 u

2)What is the most reasonable inference you could make about sulfur (S), which has four stable isotopes and an average atomic mass of 32.06 u?
A). The most common isotope of sulfur has a mass of exactly 32.06 u.
B)Approximately 6% of sulfur atoms have a mass of 32 u.
C)ll of sulfur's isotopes have a mass between 32 u and 33 u.
DThe most common isotope of sulfur is S-32.

Answer 1

Answer:

Explanation:

 .3(125)+.7(126)=37.5+88.2=125.7

2. The most common isotope is 32, because the average is very close to 32.

Answer 2

Answer:

For 1: The correct answer is Option B.

For 2: The correct answer is Option D.

Explanation:

• For 1:

Average atomic mass is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

The equation used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

We are given:

For every 100 atoms of X, 30 of them have a mass of 125.0 u and 70 of them have a mass of 126.0 u. This means that fractional abundance of X-125 isotope is 0.3 and that of X-126 isotope is 0.7

• For X-125 isotope:

Mass of the isotope = 125

Fractional abundance of isotope = 0.3

• For X-126 isotope:

Mass of the isotope = 126

Fractional abundance of isotope = 0.7

Putting values in above equation, we get:

$\text{Average atomic mass }=[(125\times 0.3)+(126\times 0.7)]=125.7u$

Hence, the correct answer is Option B.

• For 2:

We are given:

Average atomic mass of sulfur = 32.06 u

Sulfur has 4 naturally occurring isotopes: S-32, S-33, S-34 and S-36

As, the average atomic mass of sulfur lies closer to the mass of S-32 isotope. This means that the relative abundance of this isotope is the highest of all the other isotopes.

The 'S-32' isotope is the most common isotope of the given element.

Hence, the correct answer is Option D.