Question

Calculate the number of gram atoms in 100 ml mercury.
Given : Density of Hg = 13.6 $\mathrm{g}\mathrm{m}{\mathrm{l}}^{-1}$ and molar mass of Hg=$200\mathrm{g}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$
A.2.26
B.6.8
C.$6.8\times N_A$
D.${\displaystyle \frac{2.26}{N_A}}$

Answer 1

Hint: Here, first we have to calculate the mass of mercury using the formula of density, that is, $\mathrm{D}\mathrm{e}\mathrm{n}\mathrm{s}\mathrm{i}\mathrm{t}\mathrm{y}={\displaystyle \frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}}}$. Then, we have to calculate the moles of mercury. After that, using Avogadro's number we can calculate atoms present in 100 ml of mercury.

Complete step by step answer:
Given, the density of Hg is 13.6 $\mathrm{g}\mathrm{m}{\mathrm{l}}^{-1}$ and molar mass of Hg is $200\mathrm{g}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$. Now, we have to calculate the mass of mercury using the formula of density.

$\mathrm{D}\mathrm{e}\mathrm{n}\mathrm{s}\mathrm{i}\mathrm{t}\mathrm{y}={\displaystyle \frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}}}$


$\Rightarrow 13.6\mathrm{g}\mathrm{m}{\mathrm{l}}^{-1}={\displaystyle \frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{100\mathrm{m}\mathrm{l}}}$

$\Rightarrow \mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}=1360\mathrm{g}$


Therefore, the mass of mercury is 1360 g.

Now, we have to calculate the number of moles of mercury. The formula to calculate number of moles is,

$\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}={\displaystyle \frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}}}$


Now, we have to put mass of mercury (1360 g) and molar mass of mercury ($200\mathrm{g}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$) in the above equation.

$\Rightarrow \mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{s}\mathrm{o}\mathrm{f}\mathrm{m}\mathrm{e}\mathrm{r}\mathrm{c}\mathrm{u}\mathrm{r}\mathrm{y}={\displaystyle \frac{1360\mathrm{g}}{200\mathrm{g}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}}}=6.8\mathrm{m}\mathrm{o}\mathrm{l}$
measures the amount of substance present. A mole is the term that defines the number of carbon (C) atoms in 12 g of pure carbon. After so many years of experiment, it has been proved that a mole of anything contains $6.022\times {10}^{23}$ representative particles. $6.022\times {10}^{23}$ is known as Avogadro’s number or $N_A$.

1 mole of mercury contains =$6.022\times {10}^{23}$ atoms
6.8 moles of mercury contains$=6.8\times 6.022\times {10}^{23}=6.8\times N_A$ atoms.

Therefore, atoms present in 100 ml of mercury are $6.8\times N_A$.

So, the correct answer is Option C.

Note: The number $6.022\times {10}^{23}$ is named in honor of the Italian physicist Amedeo Avogadro. The Avogadro's number aids in counting very small particles. Different kinds of particles, such as molecules, atoms, ions, electrons are representative particles. One mole of anything consists of $6.022\times {10}^{23}$ representative particles.