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A metal (M^(@) = 100 gm/mole) having den...

A metal (`M^(@) = 100` gm/mole) having density 10 gm/`cm^(3)` forms a cubic lattice. If radius of an atom is 1.4 Å find packing fraction. `[N_(A)=6xx10^(23)` perm ole].

A

`0.74`

B

`0.69`

C

`0.52`

D

`0.90`

Text Solution

AI Generated Solution

To find the packing fraction of the metal, we will follow these steps: ### Step 1: Calculate the Volume of the Atom The volume \( V \) of a single atom (assuming it is spherical) can be calculated using the formula for the volume of a sphere: \[ V = \frac{4}{3} \pi r^3 \] Given that the radius \( r = 1.4 \) Å (which is \( 1.4 \times 10^{-8} \) cm), we can substitute this value into the formula. ...
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The edge length of unit cell of a metal having molecular weight 75 g mol^(-1) is 5 Å which crystallizes in cubic lattice. If the density is 2 g cc^(-1) , then find the radius of metal atom (N_(A) = 6 xx 10^(23)) . Give the answer in pm.