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Standard XII

Chemistry

introduction

The ratio of ...

Question

The ratio of isotopes of $_{17}^{37} C l$ and $_{17}^{35} C l$ in natural chlorine is 1:3. Calculate the fractional atomic mass of chlorine.

36.25

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35.5

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35.25

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36.75

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Solution

The correct option is B 35.5
Ratio of $C l^{35}$ to $C l^{37}$ is given as= $3 : 1$ Therefore, if a $4 x$ atoms are present then,
$C l^{35} = 3 x$
$C l^{37} = x$
Now, for fractional atomic mass:Atomic mass of chlorine $= \frac{(35 \times 3) + (37 \times 1)}{4} = 35.5$ $a . m . u$

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Similar questions

The two isotopes of chlorine are ³⁵₁₇Cl and ³⁷₁₇Cl in the ratio of 3:1. What is the average atomic mass of the chlorine atom?

Q. Chlorine has two isotopes,

_{17}^{35} Cl

and

_{17}^{37} Cl

. Calculate the average atomic mass of chlorine atom (in a.m.u.) based on the given information.

\begin{matrix} S.No & Isotopes & Percentage abundance (%) 1 & _{17}^{35} Cl & 75 2 & _{17}^{37} Cl & 25 \end{matrix}

Ordinary chlorine has two isotopes $_{17}^{35} C l$ and $_{17}^{37} C l$ in the ration of 3:1 Calculate the relative atomic mass of chlorine.

Q. Natural chlorine contains chlorine in the form of the isotope

^{35} C l (75.5

%) and

^{37} C l

(24.5%). Calculate the average atomic mass of natural chlorine.

Q. Assertion :The mass of chlorine is 35.5. Reason: Nature chlorine exists in two isotopic forms and these two isotopes have a natural abundance ratio of 3:1.

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The ratio of isotopes of $_{17}^{37} C l$ and $_{17}^{35} C l$ in natural chlorine is 1:3. Calculate the fractional atomic mass of chlorine.

The correct option is B 35.5
Ratio of $C l^{35}$ to $C l^{37}$ is given as= $3 : 1$ Therefore, if a $4 x$ atoms are present then,
$C l^{35} = 3 x$
$C l^{37} = x$
Now, for fractional atomic mass:Atomic mass of chlorine $= \frac{(35 \times 3) + (37 \times 1)}{4} = 35.5$ $a . m . u$

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The ratio of isotopes of 3717Cl and 3517Cl in natural chlorine is 1:3. Calculate the fractional atomic mass of chlorine.

The correct option is B 35.5Ratio of Cl35 to Cl37 is given as=3:1Therefore, if a 4x atoms are present then,Cl35=3xCl37=xNow, for fractional atomic mass:Atomic mass of chlorine=(35×3)+(37×1)4=35.5 a.m.u

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The ratio of isotopes of $_{17}^{37} C l$ and $_{17}^{35} C l$ in natural chlorine is 1:3. Calculate the fractional atomic mass of chlorine.

The correct option is B 35.5
Ratio of $C l^{35}$ to $C l^{37}$ is given as= $3 : 1$ Therefore, if a $4 x$ atoms are present then,
$C l^{35} = 3 x$
$C l^{37} = x$
Now, for fractional atomic mass:Atomic mass of chlorine $= \frac{(35 \times 3) + (37 \times 1)}{4} = 35.5$ $a . m . u$