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In the $\mathrm{HCl}$ molecule, the separation between the nuclei of two atoms is about $1.27 Å$ $\left(1 Å=10^{-10} \mathrm{~m}\right)$. Find the approximate location of the C.M. 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?
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Let $m \rightarrow$ mass of H-atom $\therefore 35.5 \mathrm{~m} \rightarrow$ mass of Cl-atom
Let C.M. is situated at a distance $x$ from H-atom.
$\therefore \quad$ Distance of C.M. from $\mathrm{Cl}$ atom $=(1.27-x) Å$
Considering the C.M. as the origin, the sum of moments of both the atoms about the origin will be zero.
$$
\therefore \quad m x-(1.27-x) 35.5 \mathrm{~m}=0
$$
[both have opposite directions]
$$
\begin{aligned}
& x=35.5(1.27-x) . \\
& x+35.5 x=35.5 \times 1.27 \\
& 36.5 x=35.5 \times 1.27 \\
\Rightarrow & x=\frac{35.5 \times 1.27}{36.5}=1.235 Å
\end{aligned}
$$
$\therefore$ The C.M. of the molecule is located at a distance of $1.235 Å$ from the $\mathrm{H}$-atom.
Let C.M. is situated at a distance $x$ from H-atom.
$\therefore \quad$ Distance of C.M. from $\mathrm{Cl}$ atom $=(1.27-x) Å$
Considering the C.M. as the origin, the sum of moments of both the atoms about the origin will be zero.
$$
\therefore \quad m x-(1.27-x) 35.5 \mathrm{~m}=0
$$
[both have opposite directions]
$$
\begin{aligned}
& x=35.5(1.27-x) . \\
& x+35.5 x=35.5 \times 1.27 \\
& 36.5 x=35.5 \times 1.27 \\
\Rightarrow & x=\frac{35.5 \times 1.27}{36.5}=1.235 Å
\end{aligned}
$$
$\therefore$ The C.M. of the molecule is located at a distance of $1.235 Å$ from the $\mathrm{H}$-atom.
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