Question

Ice crystallises in a hexagonal lattice having volume of the unit cell as \[132 \times {10^{ - 24}}\] cc. If density is 0.92 g/cc at a given temperature, then the number of water molecules per unit cell is:
A. 1
B. 2
C. 3
D. 4

Answer 1

Hint: Initially you must be aware of the formula which is used in calculating mole. You must use the method where at first you should find the mass of the unit cell and mass of water first then you should calculate the number of moles and at last find the number of molecules.

Complete step by step answer:
Given:
Volume =\[132 \times {10^{ - 24}}c{m^3}\]
\[Density{\text{ }} = 0.92\;g{\text{ }}c{m^{ - 3}}\]
1 mole of ${H_2}O$ has a mass of 18g.
Avogadro’s Number: \[6.023 \times {10^{23}}\;\]
Formula:
Mass of the unit cell = Volume × Density
Calculations:
\[Mass{\text{ }}of{\text{ }}the{\text{ }}unit{\text{ }}cell\; = \;Volume\; \times \;Density\; = 132 \times {10^{ - 24}} \times 0.92\; = 121.44 \times {10^{ - 24}}\;g\]
Molar mass of water =18.02 g mol−1

\[
Number{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ }}water\; = \dfrac{{121.44 \times {{10}^{ - 24}}}}{{18.02\;\;}}\; \\
  Number{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ }}water\; = 6.739 \times {10^{ - 24\;}}mol \\
 \]
\[\begin{array}{*{20}{l}}
  {Number{\text{ }}of{\text{ }}molecules\; = 6.739 \times {{10}^{ - 24}} \times 6.023 \times {{10}^{23}}\;molecules{\text{ }}per{\text{ }}mole\;} \\
\end{array}\]

Number of molecules = 4.05 molecules
Number of molecules = 4 molecules approximately

So, the correct answer is Option D.

Note: When the temperature is reached, the water changes to ice crystals. This will increase the concentrations of sugars and different solute gift within the combine. The accrued concentrations can depress the temperature more, and so the temperature should be lowered to make additional ice crystals. So-called "diamond dust", conjointly called ice needles or ice crystals, forms at temperatures approaching −40o C thanks to air with slightly higher wet from aloft compounding with colder, surface-based air.
\[Mass{\text{ }}of{\text{ }}building{\text{ }}block{\text{ }} = {\text{ }}variety{\text{ }}of{\text{ }}atoms{\text{ }}in{\text{ }}building{\text{ }}block{\text{ }} \times {\text{ }}mass{\text{ }}of{\text{ }}every{\text{ }}atom{\text{ }} = {\text{ }}z{\text{ }} \times {\text{ }}m\]
Where, z = variety of atoms in building block, and m = Mass of every atom