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Question:

A) Calculate the mass of a mercury atom in kg.
B) How many atoms of mercury does a drop of mercury contain, knowing that the volume of a drop of mercury is $\pu{0.05mL}$ and that the density of mercury is $13.6 \times 10^3 \mathrm{kg/m^3}$

I know that $\mathrm{1 mL = 1 cm^3}=1.0\times10^{-6} \mathrm{m^3}$

And the mass of a drop of mercury is $\mathrm{m=v}\times \mathrm{density} = 5.0\times10^{-18} \times 13.6\times{10^3} =6.8\times10^{-14}$

What I want to know is how to proceed to find how many atoms mercury are present in its one drop? Do I simply divide the mass of a drop of mercury by the mass of a mercury atom, or do I use the Avogadro's number? I don't want a straight-off answer, I only wish to know how to calculate it.

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    $\begingroup$ You've got an error in your volume in part B. You state that $1ml = 1cm^3 = 1\times 10^{-6}m^3$, but when you use the volume of a drop, 0.05ml, you somehow change that into $5\times 10^{-18}m^3$. You don't need to cube the number, the cube is there solely as part of the units. It's $5\times 10^{-8}m^3$. And yes, you just divide the mass of the drop by the mass of a single atom, assuming both masses are in the same units. $\endgroup$ – Jason Patterson Oct 7 '14 at 17:43
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You can use dimensional analysis. It is a very useful tool to solve equations such as these.

What you have is the volume of one drop, $V(ml)$, the density of mercury, $\rho(kg\cdot{}m^{-3})$, the molar mass of mercury, $Mr(g\cdot{}mol^{-1})$, and the Avogadro's constant, $N_A(mol^{-1}$).

To find out the number of atoms in a drop of mercury, all you have to do is to manipulate those variables via dimensional analysis.

You have: $\frac{Mr(g)}{(mol)},\frac{N_A(atoms)}{(mol^{})},\frac{\rho{(kg)}}{(m^3)},\frac{V(ml)}{drop}$ and you know that $\frac{1(kg)}{1000(g)},\frac{10^6(ml)}{(m^3)}$

Now, try arranging them to form an equation such that the units cancel off,in order to get the answer you need.

A: Mass of mercury:

$$\frac{kg}{atom}=\frac{1(kg)}{1000(g)}\cdot{} \frac{Mr(g)}{(mol)}\cdot{}\frac{(mol^{})}{N_A(atom)} $$

Notice that this is dimensionally equivalent. Therefore,

$$\frac{kg}{atom}=\frac{Mr}{1000\cdot{}N_A} $$

B: Atoms of mercury per drop:

$$\frac{atoms}{drop}= \frac{N_A(atoms)}{(mol^{})}\cdot{} \frac{(mol)}{Mr(g)}\cdot{} \frac{1000(g)}{1(kg)}\cdot{} \frac{\rho{(kg)}}{(m^3)}\cdot{} \frac{(m^3)}{10^6(ml)}\cdot{} \frac{V(ml)}{drop}$$

Notice that this is dimensionally equivalent. Therefore, $$\frac{atoms}{drop}= \frac{N_A\cdot{\rho}\cdot{V}}{Mr\cdot{10^3}}$$

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