Question

1. The density of mercury is $13.6gm{L^{ - 1}}$. Calculate the diameter of an atom of mercury assuming that each atom of mercury is occupying a cube of edge length equal to the diameter of mercury atom. (Atomic mass of mercury$= 200$)
2. A metal M of atomic mass $54.94$ has a density of $7.42gc{c^{ - 1}}$. Calculate the apparent volume occupied by one atom of the metal.
3. Find the charge of $1g$ ion of ${N^{3 - }}$ in coulomb.
4. Calculate the volume at NTP occupied by $6.25g$ of nitrogen.

Answer 1

As we know that one mole of any atom or molecules is equivalent to the ratio of given mass to the molecular mass and charge on one electron is $1.6 \times {10^{ - 19}}C$and at normal conditions of temperature and pressure is equivalent to $1atm$and $293K$.

Complete Step by step answer:
We know that one mole is equivalent to the ratio of given mass to the molecular mass, charge on one electron is $1.6 \times {10^{ - 19}}C$ and at normal conditions of temperature and pressure is equivalent to $1atm$ and $293K$. We also know that density is given as the mass to the volume ratio.

Following these conditions we can calculate the diameter, volume or any other parameters.
1.In the first question we are given that the density of mercury is $13.6gm{L^{ - 1}}$, atomic mass of mercury is $200g$. Now let us assume that the diameter of one mercury atom is $a$, so its volume will be equal to ${a^3}$ and volume of cube or volume of N mercury molecules will be ${a^3} \times {N_A}$. Using the formula for density of cube we get:
$\dfrac{{mass}}{{volume}} = \dfrac{{atomic\;mass}}{{{a^3} \times {N_A}}}$
$\Rightarrow 13.6 = \dfrac{{200}}{{{a^3} \times 6.022 \times {{10}^{23}}}}$
$\Rightarrow {a^3} = \dfrac{{200}}{{13.6 \times 6.022 \times {{10}^{23}}}} \\\ \Rightarrow a = 2.9 \times {10^{ - 8}}cm \\\$
As the question says that edge length is equal to the diameter therefore the diameter of mercury atom is$a = 2.9 \times {10^{ - 8}}cm$.
2. In the second question, atomic mass of the metal is $54.94g$ and density is $7.42gc{c^{ - 1}}$, we know that $1$ mole or $54.94g$ of metal M contains ${N_A}$ atoms, so we can calculate the mass using the details.
Mass of one atom $= \dfrac{{54.94}}{{6.022 \times {{10}^{23}}}}$
$mass = 9.12 \times {10^{ - 23}}g$
Now volume occupied by $1$ atom will be: $\dfrac{{mass}}{{density}}$
$volume = \dfrac{{9.12 \times {{10}^{ - 23}}}}{{7.42}} \\\ \Rightarrow volume = 1.23 \times {10^{ - 23}}cc \\\$
Thus the correct answer is $Volume = 1.23 \times {10^{ - 23}}cc$.
3. We know that charge on one mole of electron is $96500\;C$, so charge on ${N^{3 - }}$ will be $3 \times 96500\;C$because ${N^{3 - }}$ contains three electrons.
So charge in $1g$ ion means charge in $1$ mole.
So charge on $1g$ of nitrogen ion $= 3 \times 96500 \times 1$
Charge$= 2.89 \times {10^5}C$.
Therefore the correct answer is charge$= 2.89 \times {10^5}C$.
4. In the last question we are given the condition of NTP where the pressure is equal to $1atm$ and temperature is equal to $293K$. By applying the ideal gas equation formula we get:
$PV = nRT \\\ \Rightarrow V = \dfrac{{nRT}}{P} \\\ \Rightarrow V = \dfrac{{6.25 \times 0.0821 \times 293}}{{28 \times 1}} \\\ \Rightarrow V = 5.37L \\\$
Where n is the number of moles of nitrogen given by mass to atomic mass ratio.
Therefore the correct answer is $Volume = 5.37L$.

Note: Just like NTP, at standard condition of temperature and pressure we can calculate the volume by first calculating the number of moles and then multiplying these moles with $22.4L$ which is the condition of volume at STP.