Given the mean atomic mass of chlorine prepared in the laboratory to be 35.5 and assuming that chlorine contains two isotopes of mass number 35 nad 37, what is the percentage composition of the isotope of mass number 35?
20
25
50
75
90
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The answer is 75
To know the logic is that when given a smaller mass the relative abundance will be higher than the relative abundance of a larger mass
solution
35.5=35∞, + 37∞,, (note that ∞, and ∞,, are the relative abundance)
but, ∞,+∞,,=1
∞,=1-∞,,
35.5=35(1-∞,,)+37∞,,
35.5=35-35∞,,+37∞,,
35.5-35=-35∞,+37∞,,
0.5=2∞,,
divide both sides by 2
∞,,=0.25×100
∞,,=25%
but ∞,=1-∞,,
∞,=1-0.25
∞,=0.75×100
∞,=75%
therefore it means that the % value for mass 35 is 75 and for 37 is 25
To determine the percentage composition of the isotope with mass number 35, we use the formula for the weighted average atomic mass:
\text{Mean atomic mass} = (f_1 \times m_1) + (f_2 \times m_2)
where:
, (the mass numbers of the isotopes)
and are the fractional abundances of the isotopes ()
The given mean atomic mass of chlorine is 35.5
Step 1: Express in terms of
Let be the fraction of the isotope with mass 35. Then:
f_2 = 1 - f_1
Substituting values into the equation:
35.5 = (f_1 \times 35) + ((1 - f_1) \times 37)
Step 2: Solve for
Expanding the equation:
35.5 = 35f_1 + 37 - 37f_1
35.5 = 37 - 2f_1
Rearrange:
2f_1 = 37 - 35.5
2f_1 = 1.5
f_1 = \frac{1.5}{2} = 0.75
Step 3: Convert to Percentage
\text{Percentage of isotope 35} = 0.75 \times 100 = 75\%
Answer:
D. 75%
The atomic mass =35.5-35=0.05 and the difference in mass of isotope 37=37-35.5=1.5
sum of ratio=0.05+1.5=2.0
%composition=0.5× 100/2=25%
