Chlorine exists in two isotopic forms, $C l - 37$ and $C l - 35$ but its atomic mass is $35.5$ . This indicates the ratio of $C l - 37$ and $C l - 35$ is approximately:

1 : 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 : 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 : 3

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3 : 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $1 : 3$
Let relative abundance of Cl-37 be $A_{1}$ % with mass $M_{1} = 37$ amu

$∴$ The relative abundance of Cl-35 = $A_{2}$ % with mass, $M_{2} = 35$ amu

where $A_{1} + A_{2} = 100$

Thus $A_{2}$ %= $(100 - A_{1})$ %

Using average atomic mass, $M_{a v g} = \frac{M_{1} A_{1} + M_{2} A_{2}}{A_{1} + A_{2}}$

$M_{a v g} = \frac{37 \times A_{1} + 35 \times (100 - A_{1})}{A_{1} + (100 - A_{1})} = \frac{2 A_{1} + 3500}{100}$

Given that: $M_{a v g} = 35.5$

$∴ \frac{2 A_{1} + 3500}{100} = 35.5$

$2 A_{1} = 35.5 \times 100 - 3500 = 50$

$A_{1} = \frac{50}{2} = 25$ and $A_{2} = 100 - A_{1} = 100 - 25$ =75

Thus the ratio of isotope abundance of $C l - 37 t o C l - 35$ is $A_{1} : A_{2} = 25 : 75 = 1 : 3$

So, the correct option is $C$

Suggest Corrections

Similar questions

Question 13

Chlorine exists in two isotopic forms. Cl - 37 and Cl - 35 but its atomic mass is 35.5. This indicates the ratio of Cl - 37 and Cl - 35 is approximately

(a) 1 : 2 (b) 1 : 1 (c) 1 : 3 (d) 3 : 1

Naturally occurring chlorine consists of two isotopes whose atomic weights are 35 and 37. The atomic weight of natural chlorine is 35.5, then what is the ratio of $C l^{35} t o C l^{37}$ ?