1.

Tungsten has bcc lattice. Each edge of the unit cell is 316 pm and the density of the metal is 19.35g cm^(-3). How many atoms are present in 50 g of this element?

Answer»


Solution : Z=2 for BCC lattice, a=316 pm, d = 19.35 g `cm^(-3)`, M=50g
Volume of 50 g tungsten = `("Mass")/("Density") =(50)/(19.35)=2.58 cm^3`
Volume of UNIT cell = `a^3 (316 xx 10^(-10))^3 cm^3 = 3.15 xx 10^(-23) cm^3`
Number of unit CELLS = `("Volume of element")/("Volume of unit cell")=(2.58)/(3.15 xx 10^(-23)) = 8.19 xx 10^(22)`
`:.` Number of ATOMS in 50 g element = `2 xx 8.19 xx 10^(22)= 1.638 xx 10^(23)` atoms = `1.64 xx 10^(23)` atoms


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