How many unit cells are present in a cube­-shaped deal crystal of NaCl of mass 1.00 g

[ Atomic masses : Na =23,Cl=35.5]

  • Option 1)

    2.57\times 10^{21}

  • Option 2)

    5.14\times 10^{21}

  • Option 3)

    1.28\times 10^{21}

  • Option 4)

    1.71\times 10^{21}

 

Answers (1)

As we learnt in

Density of cubic unit cell -

d = \frac{zM}{a^3N_o}

 

- wherein

Where,

d = density of crystal

z = no. of effective constituent particles in one unit cell

M = molecular weight

a = edge length of unit cell

No = 6.022*1023

 

 

 

mass(m)= density \times × volume= 1.00g

Mol. wt.(M)of NaCl= 23 + 35.5 = 58.5

Number\: of \: unit \: cell\: present \: in\: a\: cube \: shaped \: crystal \:o\! f\: NaCl \: o\! f\: mass\: 1.00 g =\frac{\rho \times a^{3}\times N_{A}}{M\times Z}= \frac{m\times N_{A}}{M\times Z}

                                                    = \frac{1\times 6.023\times 10^{23}}{58.5\times 4}

( In NaCl each unit cell has 4 NaCl units. Hence z = 4)

number\: of\: unit \: cells\: = 0.02573 \times 10^{23}

                                                = 2.57 \times 10 ^{21}

Correct option is 1.


Option 1)

2.57\times 10^{21}

This is the correct option.

Option 2)

5.14\times 10^{21}

This is an incorrect option.

Option 3)

1.28\times 10^{21}

This is an incorrect option.

Option 4)

1.71\times 10^{21}

This is an incorrect option.

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