# How many atoms of mercury in a volume of 62.5*cm^3?

Sep 3, 2017

Well, we know that $\text{density,}$ $\rho = \text{Mass"/"Volume}$

#### Explanation:

And thus $\text{Mass"=rhoxx"volume}$

=13.534*g*cm^-3xx62.5*cm^3=??*g

And $\text{moles of mercury"="mass"/"molar mass}$

=(13.534*g*cm^-3xx62.5*cm^3)/(200.59*g*mol^-1)=??*mol

And we know that there are $6.022 \times {10}^{23}$ individual items of stuff in a mole....

And so $\text{atoms of mercury}$

$= \frac{13.534 \cdot g \cdot c {m}^{-} 3 \times 62.5 \cdot c {m}^{3}}{200.59 \cdot g \cdot m o {l}^{-} 1} \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$

=??