In the $HCl$ molecule,the separation between the nuclei of the two atoms is about $1.27 \; \mathring{A} \; (1 \; \mathring{A} = 10^{-10} \; m)$. Find the approximate location of the center of mass $(CM)$ of the molecule,given that a chlorine atom is about $35.5$ times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

(N/A) Let the position of the $H$ atom be at $x_H = 0$ and the position of the $Cl$ atom be at $x_{Cl} = 1.27 \; \mathring{A}$.
Let $m_H = m$ be the mass of the hydrogen atom.
Then,the mass of the chlorine atom is $m_{Cl} = 35.5 \; m$.
The center of mass $(X_{CM})$ is given by the formula:
$X_{CM} = \frac{m_H x_H + m_{Cl} x_{Cl}}{m_H + m_{Cl}}$
Substituting the values:
$X_{CM} = \frac{m(0) + (35.5 \; m)(1.27 \; \mathring{A})}{m + 35.5 \; m}$
$X_{CM} = \frac{35.5 \; m \times 1.27 \; \mathring{A}}{36.5 \; m}$
$X_{CM} = \frac{35.5 \times 1.27}{36.5} \; \mathring{A} \approx 1.235 \; \mathring{A}$
This distance is measured from the hydrogen atom.
Alternatively,the distance from the chlorine atom is $1.27 \; \mathring{A} - 1.235 \; \mathring{A} = 0.035 \; \mathring{A}$ (approximately $0.037 \; \mathring{A}$ depending on rounding).
Thus,the center of mass is located at approximately $0.037 \; \mathring{A}$ from the chlorine atom.