Question

Ice crystallizes in a hexagonal lattice. At the low temperatures, the lattice constants were $a{\text{ = 4}}{\text{.53}}\,\mathop {\text{A}}\limits^{\text{o}} $ and $b{\text{ = 7}}{\text{.41}}\,\mathop {\text{A}}\limits^{\text{o}} $. How many ${H_2}O$ molecules are contained in a unit cell [d (ice)=$0.92\,g/c{m^3}$]?
$
  A)\,4 \\
  B)\,3 \\
  C)\,2 \\
  D)\,1 \\
 $

Answer 1

Hint: We have the sides given, so we have to calculate the volume. From the volume and density, we can calculate the mass, and from the obtained mass, we can calculate moles. The calculated moles are converted into molecules by using Avogadro’s number.

Formula used: For calculating the volume, we can use the formula,
$V = {a^2}b\,\sin \,{60^ \circ }$
Here, a and b are the sides of the hexagonal lattice
For calculating the mass, we can use the formula,
${\text{Mass = Volume X Density}}$
For converting grams to moles, we can use the formula,
${\text{Moles = }}\dfrac{{{\text{Grams}}}}{{{\text{Molecular}}\,{\text{mass}}}}$
For converting moles to molecules, we use the formula,
${\text{Molecules = Moles X Avogadro's}}\,{\text{number}}$


Complete step by step answer:
In the given data, we have the sides of the hexagonal lattice and density of the ice.
The given data is,
Side of the volume (a) is $4.53 \times {10^{ - 8}}\,cm$
Side of the volume (b) is $7.41 \times {10^{ - 8}}\,cm$
Density of ice is $0.92\,g/c{m^3}$
We can calculate the volume of the lattice using sides of the hexagonal lattice.
$Volume = {a^2}b\sin {60^ \circ }$
We know the value of a and b. so we can substitute the value here,
$
  Volume = {\left( {4.53 \times {{10}^{ - 8}}cm} \right)^2}\left( {7.41 \times {{10}^{ - 8}}\,cm} \right)\sin {60^ \circ } \\
  Volume = 1.317 \times {10^{ - 22}}\,c{m^3} \\
 $
The calculated volume is $1.317 \times {10^{ - 22}}\,c{m^3}$
We know that density is mass per unit volume. From the volume of the lattice, we can now calculate the mass using volume and density.
$
  Mass = Volume \times Density \\
  Mass = \left( {1.317 \times {{10}^{ - 22}}\,c{m^3}} \right)\left( {0.92\,g/c{m^3}} \right) \\
  Mass = 1.2116 \times {10^{ - 22}}\,g \\
 $
The calculated mass is $1.2116 \times {10^{ - 22}}\,g$
From the obtained mass, we calculate the number of moles using the molecular mass.
We can calculate the number of moles by dividing grams to the molecular mass.
The molecular mass of water is $18\,g/mol$
$
  Moles = \dfrac{{Grams}}{{Molecular\,mass}} \\
  Moles = \dfrac{{1.2116 \times {{10}^{ - 22}}\,g}}{{18\,g/mol}} \\
  Moles = 6.7311 \times {10^{ - 24}}\,mol \\
 $
We can see the calculated number of moles is $6.7311 \times {10^{ - 14}}\,mol$
We can calculate the number of molecules using Avogadro's number.
We can obtain the number of molecules by multiplying the moles with the Avogadro’s number.
The value of Avogadro’s number is $6.023 \times {10^{23}}$
$
  Molecules = Moles \times Avogadro's\,number \\
  Molecules = 6.7311 \times {10^{ - 24}}\,moles \times 6.023 \times {10^{23}}\,molecules\,per\,mol \\
  Molecules = 4.06\,\,molecules \\
 $
The number of molecules per unit cell is ${\text{4}}{\text{.}}$

$\therefore $Option (A) is correct.

Note:
We can also calculate the number of molecules per unit using the formula,
$Z = \dfrac{{\rho V{N_A}}}{M}$.
 In the formula, $\rho $ is the density, V is the volume, ${N_A}$ is the Avogadro’s number, M is the molecular mass and Z is the number of molecules per unit cell.