isotopes of chlorine with mass number 35 and 37 exist in ratio
Answers
The ratio in which two isotopes of chlorine with mass no 35 and 37 exist is 3: 1 i. e. (75 : 25) %.
Let the percentage existence of Cl 35 be x . Then percentage existence of Cl 37 is 100 - x.
Now ,
✔️Avg molecular weight be M .
✔️Molecular weight of isotope 1 be M1
✔️Molecular weight of isotope 2 be
M2.
✔️Relative abundance of isotope1 be
x.
✔️Relative abundance of isotope 2 be
100 -x.
Then,
➡️ M = [ M1 x + M2 (100-x) ] / 100
35.5 × 100 = 35x + 37 (100 - x )
3550 = 35x + 3700 - 37x
2x = 150
✔️x = 75
✔️100 - x = 100 - 75 = 25
Hence,
✔️Relative abundance of Cl 35 is 75%
✔️Relative abundance of Cl 37 is 25%
➡️Ratio of Cl 35 to Cl 37 = 75% : 25%
➡️Ratio = 3 : 1
Hope it helps....
Answer:
Isotopes of Chlorine with mass number 35 and mass number 37 exist in the ratio 3 : 1
Explanation:
The molar mass of Chlorine is 35.5
Let the mole fraction of isotope 35 be x and that of isotope 37 be (1 - x).
To get the molar mass of Chlorine given the isotopes we have :
Molar mass = Sum of (Relative abundance × mole fraction)
Let's substitute these values in the formula above.
We have molar mass = 35.5
Therefore :
35.5 = 37 × (1 - x) + 35 × x
35.5 = 37 - 37x + 35x
35.5 - 37 = - 2x
-1.5 = - 2x
x = - 1.5/-2
x = 0.75
The percentage composition of isotope 35 is = 0.75 × 100% = 75%
That of isotope 37 = 100% - 75% = 25%
From this we can get their ratio of existence.
The ratio is therefore given as :
75% : 25%
75/25 = 3 : 1
Their ratio of existence is therefore given by :
3 : 1