Ordinary chlorine has 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.

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Solution

$_{17}^{35} C l$ = 75%

$_{17}^{37} C l$ = 25%

relative atomic mass of Cl=35 x 75/100 + 37 x 25/100 = 35.5%

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Similar questions

_{17}^{35} Cl

and

_{17}^{37} Cl

are two isotopes of Chlorine. If atomic mass of Chlorine is

35.5

, then the mass ratio of these isotopes is

x : 1

, where

x

is :

Chlorine has two isotopes having atomic masses of 35u and 37u each. They are present in the ratio 3:1 respectively. The average atomic mass of Chlorine is _______u.

Q. Chlorine has two isotopes of atomic mass units 34.97 and 36.97. The relative abundance of an isotope is 0.755 and 0.245 respectively. Find the average atomic mass of chlorine.

Q. Chlorine has two stable isotopes: Cl-35 and Cl-37 with atomic masses 34.96 and 36.95, respectively. If the average mass of chlorine is 35.43, calculate the percentage abundance.

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}

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Ordinary chlorine has has two isotopes3517CL and 1737CL in the ration of 3:1 Calculate the relative atomic mass of chlorine.

3517Cl = 75% 3717Cl = 25% relative atomic mass of Cl=35 x 75/100 + 37 x 25/100 = 35.5%

Ordinary chlorine has 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.

$_{17}^{35} C l$ = 75%

$_{17}^{37} C l$ = 25%

relative atomic mass of Cl=35 x 75/100 + 37 x 25/100 = 35.5%