
If the atomic mass of \[{\text{Li}}\] and \[{\text{Cl}}\] are 7 and \[{\text{35}}{\text{.5}}\] amu respectively, then what would be the number of \[{\text{LiCl}}\] molecule in cube of \[{\text{LiCl}}\] with \[{\text{length}} \times {\text{breath}} \times {\text{height}} = 1 \times 1 \times 1{\text{ c}}{{\text{m}}^3}\] ?
A.Nearly \[2.5 \times {10^{22}}\]
B.Nearly \[4 \times {10^{20}}\]
C.Nearly \[12 \times {10^{30}}\]
D.Nearly \[14 \times {10^{23}}\]
Answer
584.4k+ views
Hint: Using the molar mass we will first calculate the molecular mass. Then the number of molecules can be calculated using the formula by substituting the known variable. \[{\text{LiCl}}\] forms a BCC cubic lattice and hence, the number of atoms in one unit will be 2.
Formula used:
\[{\text{number of molecules in unit cube }} = \dfrac{{{\text{Z}} \times {{\text{N}}_{\text{A}}}}}{{{\text{Molar mass}}}}\]
Here Z is the number of atoms in a single unit cell and \[{{\text{N}}_{\text{A}}}\] is Avogadro’s constant.
Complete step by step solution:
\[{\text{LiCl}}\] forms a BCC unit cell, that is, a body centred unit cell. In Bcc unit cell atoms are present at the cornered as well as at the body centre of the cube. The contribution of the atom at the corner is one eighth and the number of atoms in the corner are 8. There is one atom present in the body centre and the contribution of atoms present in the body centre is 1. Hence the total number of atoms in the \[{\text{LiCl}}\] unite cell will be:
\[\dfrac{1}{8} \times 8 + 1 \times 1 = 2\]
The molar mass of \[{\text{LiCl}}\] will be: \[35.5 + 7 = 42.5\]
The value of Avogadro constant is fixed and that is: \[6.022 \times {10^{23}}\]
Putting the above variables in the formula we will get:
\[{\text{number of molecules in unit cube }} = \dfrac{{2 \times 6.022 \times {{10}^{23}}}}{{{\text{42}}{\text{.5}}}} \simeq 2.5 \times {10^{22}}\]
Hence, the correct option is A.
Note:
Just like a cell for the living organism, a unit cell also acts as a building block for the complete lattice. The unit cell follows the same properties as that of the whole lattice. A unit cell is an array of points joined together in 3 D space to form lattice. There are various types of unit cell such as primitive or simple cubic unit cell, face centred unit cell, end centred unit cell and edge centred unit cell.
Formula used:
\[{\text{number of molecules in unit cube }} = \dfrac{{{\text{Z}} \times {{\text{N}}_{\text{A}}}}}{{{\text{Molar mass}}}}\]
Here Z is the number of atoms in a single unit cell and \[{{\text{N}}_{\text{A}}}\] is Avogadro’s constant.
Complete step by step solution:
\[{\text{LiCl}}\] forms a BCC unit cell, that is, a body centred unit cell. In Bcc unit cell atoms are present at the cornered as well as at the body centre of the cube. The contribution of the atom at the corner is one eighth and the number of atoms in the corner are 8. There is one atom present in the body centre and the contribution of atoms present in the body centre is 1. Hence the total number of atoms in the \[{\text{LiCl}}\] unite cell will be:
\[\dfrac{1}{8} \times 8 + 1 \times 1 = 2\]
The molar mass of \[{\text{LiCl}}\] will be: \[35.5 + 7 = 42.5\]
The value of Avogadro constant is fixed and that is: \[6.022 \times {10^{23}}\]
Putting the above variables in the formula we will get:
\[{\text{number of molecules in unit cube }} = \dfrac{{2 \times 6.022 \times {{10}^{23}}}}{{{\text{42}}{\text{.5}}}} \simeq 2.5 \times {10^{22}}\]
Hence, the correct option is A.
Note:
Just like a cell for the living organism, a unit cell also acts as a building block for the complete lattice. The unit cell follows the same properties as that of the whole lattice. A unit cell is an array of points joined together in 3 D space to form lattice. There are various types of unit cell such as primitive or simple cubic unit cell, face centred unit cell, end centred unit cell and edge centred unit cell.
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