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प्रश्न
Derive the relationship between density of substance, its molar mass, and the unit cell edge length. Explain how you will calculate the number of particles, and a number of unit cells in x g of metal.
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उत्तर
- Relationship between density of a substance, its molar mass and the unit cell edge length:
- If edge length of cubic unit cell is ‘a’, then the volume of unit cell is a3.
- Suppose that mass of one particle is ‘m’ and that there are ‘n’ particles per unit cell.
∴ Mass of unit cell = m × n .....(1) - The density of unit cell (ρ), which is same as density of the substance is given by:
`rho = "Mass of unit cell"/"Volume of a unit cell"`
= `("m" xx "n")/"a"^3` = Density of substance …(2) - Molar mass (M) of the substance is given by:
M = mass of one particle × number of particles per mole
= m × NA (NA is Avogadro number)
Therefore, m = `"M"/"N"_"A"` .....(3) - Combining equations (2) and (3), gives
`rho = ("n M")/("a"^3 "N"_"A")` .....(4)
- The density (ρ) and molar mass (M) of metal are related to each other through unit cell parameters as given below:
`rho = "n"/"a"^3 xx "M"/"N"_"A"`
∴ M = ρ`("a"^3 "N"_"A")/"n"`
where, ‘n’ is the number of particles in unit cell and ‘a3’ is the volume of unit cell.
- The number of particles in x g of metallic crystal:
∴ Molar mass, M, contains NA particles.
∴ x g of metal contains `(x"N"_"A")/"M"` particles.
Substitution of M in the above equation gives
Number of particles in ‘x’ g = `(x"N"_"A")/(rho"a"^3"N"_"A"//"n") = "x n"/(rho "a"^3)` - Number of unit cells in x g of metallic crystal:
‘n’ particles correspond to 1 unit cell.
∴ `(x"n")/(rho"a"^3)` particles correspond to `(x"n")/(rho"a"^3) xx 1/"n"` unit cells.
∴ Number of unit cells in ‘x’ g metal = `x/(rho"a"^3)`
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