Answer
Verified
428.1k+ views
Hint: Here, first we have to calculate the mass of mercury using the formula of density, that is, ${\rm{Density}} = \dfrac{{{\rm{Mass}}}}{{{\rm{Volume}}}}$. 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 ${\rm{g}}\,{\rm{m}}{{\rm{l}}^{ - 1}}$ and molar mass of Hg is $200\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$. Now, we have to calculate the mass of mercury using the formula of density.
${\rm{Density}} = \dfrac{{{\rm{Mass}}}}{{{\rm{Volume}}}}$
$ \Rightarrow 13.6\,{\rm{g}}\,{\rm{m}}{{\rm{l}}^{ - 1}} = \dfrac{{{\rm{Mass}}}}{{100\,{\rm{ml}}}}$
$ \Rightarrow {\rm{Mass}} = 1360\,{\rm{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,
${\rm{Number}}\,{\rm{of}}\,{\rm{moles = }}\dfrac{{{\rm{Mass}}}}{{{\rm{Molar}}\,{\rm{mass}}}}$
Now, we have to put mass of mercury (1360 g) and molar mass of mercury ($200\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$) in the above equation.
$ \Rightarrow {\rm{Number}}\,{\rm{of}}\,{\rm{moles}}\,{\rm{of}}\,{\rm{mercury}} = \dfrac{{1360\,{\rm{g}}}}{{200\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}}} = 6.8\,{\rm{mol}}$
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.
Complete step by step answer:
Given, the density of Hg is 13.6 ${\rm{g}}\,{\rm{m}}{{\rm{l}}^{ - 1}}$ and molar mass of Hg is $200\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$. Now, we have to calculate the mass of mercury using the formula of density.
${\rm{Density}} = \dfrac{{{\rm{Mass}}}}{{{\rm{Volume}}}}$
$ \Rightarrow 13.6\,{\rm{g}}\,{\rm{m}}{{\rm{l}}^{ - 1}} = \dfrac{{{\rm{Mass}}}}{{100\,{\rm{ml}}}}$
$ \Rightarrow {\rm{Mass}} = 1360\,{\rm{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,
${\rm{Number}}\,{\rm{of}}\,{\rm{moles = }}\dfrac{{{\rm{Mass}}}}{{{\rm{Molar}}\,{\rm{mass}}}}$
Now, we have to put mass of mercury (1360 g) and molar mass of mercury ($200\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$) in the above equation.
$ \Rightarrow {\rm{Number}}\,{\rm{of}}\,{\rm{moles}}\,{\rm{of}}\,{\rm{mercury}} = \dfrac{{1360\,{\rm{g}}}}{{200\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}}} = 6.8\,{\rm{mol}}$
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
10 examples of friction in our daily life
When people say No pun intended what does that mea class 8 english CBSE