# What is the mass of 4.49*10^25 particles of HCl?

Mar 8, 2016

$\text{Number of HCl particles"/"Number of HCl particles per mol}$ $\times 36.5 \cdot g \cdot m o {l}^{-} 1$ $=$ ??
It is a fact that $6.022 \times {10}^{23}$ individual $H C l$ particles have a mass of $36.5$ $g$. This is simply another way of saying that the molar mass of $H C l$, the mass of $\text{Avogadro's Number}$ of $H C l$ particles, is $36.5 \cdot g \cdot m o {l}^{-} 1$.
$\text{Number of HCl particles"/"Number of HCl particles per mol}$ $\times 36.5 \cdot g \cdot m o {l}^{-} 1$ $=$ $\frac{4.49 \times {10}^{25}}{6.022 \times {10}^{23}}$ $\times$ $36.5 \cdot g$ $=$ ?? $g$