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  • Tungsten has a density of 19.35 gcm-3 and the length of the side of the unit cell is 316 pm. The unit cell in the most important crystalline form of tungsten is the body-centered cu

Tungsten has a density of 19.35 gcm-3 and the length of the side of the unit cell is 316 pm. The unit cell in the most important crystalline form of tungsten is the body-centered cubic unit cell. How many atoms of the element does 50g of the element contain? 

Option: 1

1.61\times10^{23}  atoms


Option: 2

16.1\times10^{23} atoms


Option: 3

13.4\times10^{23} atoms


Option: 4

1.98\times10^{23} atoms


Answers (1)

best_answer

Density of Lattice Matter (d)
It is the ratio of mass per unit cell to the total volume of a unit cell and it is found out as follows.

\mathrm{d}=\frac{\mathrm{Z} \times \text { Atomic weight }}{\mathrm{N}_{0} \times \text { Volume of unit cell }\left(\mathrm{a}^{3}\right)}
Here, d = Density
         Z = Number of atoms
         N0 = Avogadro number
         a3 = Volume
         a = Edge length
Here in order to find density of unit cell in cm3, m must be taken in g/mole and should be in cm.

-

We have given: 
Length of edge of unit cell =316 pm or 316\times10^{-10}? cm. Thus, volume of unit cell =(316\times10^{-10})^3=3.2\times10^{-23}\: cm^3/unit\: cell 

Now,

\\\mathrm{Density\: =\: \frac{Atomic\: mass\: x\: Z}{Unit\: cell\: volume\: x\: N_{0}}}\\\\\mathrm{Thus,\: Atomic\: mass\: =\: \frac{Density\: x\: Unit\: cell\: volume\: x\: N_{0}}{Z}}\\\\\mathrm{=\: \frac{19.35\: x\: 3.2\: x\: 10^{-23}\: x\: 6.023\: x\: 10^{23}}{2}\: =\: 186.5g/mol}

Now, since 186.5g of the element contains 6.023\times10^{23}? atoms

Thus, 50 g of the element contains =6.023\times10^{23}\times\frac{50}{186.5}=1.61\times 10^{23}\: atoms

Therefore, Option(1) is correct

Posted by

seema garhwal

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