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

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Answer 1 · 2017-09-03T12:20:57

Well, we know that $"density,"$ $rho="Mass"/"Volume"$

And thus $"Mass"=rhoxx"volume"$

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

And $"moles of mercury"="mass"/"molar mass"$

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

And we know that there are $6.022xx10^23$ individual items of stuff in a mole....

And so $"atoms of mercury"$

$=(13.534*g*cm^-3xx62.5*cm^3)/(200.59*g*mol^-1)xx6.022xx10^23*mol^-1$

$=??$

How many atoms of mercury in a volume of 62.5*cm^362.5*cm^3?