The isotopes of chlorine with mass numbers $35$ and $37$ whose average mass is 35.5 exist in the ratio of:

1 : 1

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3 : 1

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3 : 4

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3 : 2

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Solution

The correct option is B $3 : 1$
Let the fraction of $C l^{35}$ isomer be $p$ . The fraction of $C l^{37}$ isomer will be $1 - p$ .
Average atomic mass $= p \times m_{1} + (1 - p) m_{2}$
$∴ 35.5 = p \times 35 + (1 - p) \times 37$
So, $p = 0.75$
Hence, the ratio of two isotopes will be $p : (1 - p) = 0.75 : (1 - 0.75) = 0.75 : 0.25 = 3 : 1$

Option B is correct.

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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 $C l^{35} t o C l^{37}$ ?

^{35} C l

and

^{37} C l

are the two isotopes of chlorine, in the ratio

3 : 1

respectively. If the isotope ratio is reversed the average atomic mass of chlorine will be:

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The isotopes of chlorine with mass numbers 35 and 37 whose average mass is 35.5 exist in the ratio of:

The correct option is B 3:1Let the fraction of Cl35 isomer be p. The fraction of Cl37 isomer will be 1−p.Average atomic mass =p×m1+(1−p)m2∴35.5=p×35+(1−p)×37So, p=0.75Hence, the ratio of two isotopes will be p:(1−p)=0.75:(1−0.75)=0.75:0.25=3:1Option B is correct.

The isotopes of chlorine with mass numbers $35$ and $37$ whose average mass is 35.5 exist in the ratio of: