$_{17}^{35} Cl$ 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 :

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Solution

Average atomic mass $= \frac{A_{1} X_{1} + A_{2} X_{2}}{x_{1} + x_{2}}$
$= \frac{35 x + 37}{x + 1} = 35.5$
Thus, $x = 3$ .

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Similar questions

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.

_{17}^{35} C l

and

_{17}^{37} C l

are two isotopes of chlorine. If average atomic mass is

35.5

then ratio of these two isotopes is:

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.

Q. Atomic mass of chlorine is 35.5. It has two isotopes of atomic mass 35 and 37. The percentage of heavier isotope 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.

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3517Cl and 3717Cl 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 :

Average atomic mass=A1X1+A2X2x1+x2 =35x+37x+1=35.5Thus, x=3.

$_{17}^{35} Cl$ 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 :

Average atomic mass $= \frac{A_{1} X_{1} + A_{2} X_{2}}{x_{1} + x_{2}}$
$= \frac{35 x + 37}{x + 1} = 35.5$
Thus, $x = 3$ .