Question

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

• A. 36.25
• B. 35.5
• C. 35.25
• D. 36.75

Answer 1

The correct option is B 35.5

Ratio of $C{l}^{35}$ to $C{l}^{37}$ is given as=$3:1$Therefore, if a $4x$ atoms are present then,

$C{l}^{35}=3x$

$C{l}^{37}=x$

Now, for fractional atomic mass:Atomic mass of chlorine$=\frac{(35\times 3)+(37\times 1)}{4}=35.5$ $a.m.u$