Chlorine exists in two isotopic forms ^{35}Cl | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the percentage abundance of $^{35}Cl$ be $x$ and that of $^{37}Cl$ be $(100 - x)$.Average atomic mass is given by the formula:$	ext{Average a

Vedclass · Accepted Answer

$75 \%, 25 \%$

Vedclass · Answer

(B) Let the percentage abundance of $^{35}Cl$ be $x$ and that of $^{37}Cl$ be $(100 - x)$.Average atomic mass is given by the formula:$	ext{Average atomic mass} = \frac{(	ext{mass of } ^{35}Cl 	imes x) + (	ext{mass of } ^{37}Cl 	imes (100 - x))}{100}$Substituting the values:$35.5 = \frac{35x + 37(100 - x)}{100}$$3550 = 35x + 3700 - 37x$$3550 = 3700 - 2x$$2x = 3700 - 3550$$2x = 150$$x = 75$Therefore,the abundance of $^{35}Cl$ is $75 \%$ and the abundance of $^{37}Cl$ is $100 - 75 = 25 \%$.

Chlorine exists in two isotopic forms $^{35}Cl$ and $^{37}Cl$. If the average atomic mass of chlorine is $35.5$,what is the percentage abundance of these isotopes respectively?

Similar Questions

If $M_1$ and $M_2$ are the molecular weights of $Al(OH)_3$ and $H_3PO_3$ respectively,find their equivalent weights in the following reaction:
$Al(OH)_3 + 2H_3PO_3 \to Al(OH)(H_2PO_3)_2 + 2H_2O$

Equivalent weights of $X_2Y$ and $X_2Y_3$ are $38$ and $18$ respectively. Find the atomic masses of $X$ and $Y$:-

What is meant by $1 \ amu$ or $1 \ u$?

$A$ metal oxide is reduced by heating it in a stream of hydrogen. It is found that after complete reduction,$3.15 \, g$ of the oxide yielded $1.05 \, g$ of the metal. We may deduce that:

$4 \ g$ of copper was dissolved in concentrated nitric acid. The copper nitrate solution on strong heating gave $5 \ g$ of its oxide. The equivalent weight of copper is

Vedclass Test Series

Exam Paper Generator

Online Exam Module