(B) Let the natural abundance of ${}^{35}Cl$ be $x$ and that of ${}^{37}Cl$ be $y$.The average atomic mass is given by the formula: $M_{av} = \frac{M_

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(B) Let the natural abundance of ${}^{35}Cl$ be $x$ and that of ${}^{37}Cl$ be $y$.The average atomic mass is given by the formula: $M_{av} = \frac{M_{1}x + M_{2}y}{x + y}$.Substituting the given values: $35.45 = \frac{35x + 37y}{x + y}$.$35.45(x + y) = 35x + 37y$.$35.45x + 35.45y = 35x + 37y$.$0.45x = 1.55y$.$\frac{x}{y} = \frac{1.55}{0.45} = \frac{155}{45} \approx 3.44$.The ratio is closest to $3: 1$.

Class 11 Chemistry Some Basic Concepts of Chemistry Atomic, Molecular and Equivalent masses

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Chlorine has two naturally occurring isotopes,${}^{35}Cl$ and ${}^{37}Cl$. If the atomic mass of $Cl$ is $35.45$,the ratio of natural abundance of ${}^{35}Cl$ and ${}^{37}Cl$ is closest to (in $: 1$)

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Some Basic Concepts of Chemistry

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Atomic, Molecular and Equivalent masses

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Chlorine has two naturally occurring isotopes,${}^{35}Cl$ and ${}^{37}Cl$. If the atomic mass of $Cl$ is $35.45$,the ratio of natural abundance of ${}^{35}Cl$ and ${}^{37}Cl$ is closest to (in $: 1$)

Similar Questions

$71 \ g$ of chlorine combines with a metal to form $111 \ g$ of its chloride. The chloride is isomorphous with $MgCl_2 \cdot 6H_2O$. The atomic weight of the metal is:

Sulphur forms the chlorides $S_2Cl_2$ and $SCl_2$. The equivalent mass of sulphur in $SCl_2$ is $...$ $g/mol$.

An appropriate method for molecular weight determination of chloroform $(CHCl_3)$ is:

$NaOH + H_3PO_4 \to NaH_2PO_4 + H_2O$; What is the equivalent weight of phosphoric acid in this reaction?

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