Advertisements
Advertisements
प्रश्न
An element crystallises in a b.c.c lattice with cell edge of 500 pm. The density of the element is 7.5g cm-3. How many atoms are present in 300 g of the element?
Advertisements
उत्तर
a = 500 pm = 500 x 10-10 cm
z = 2
m = 300 g
`Density(d)=(zM)/(a^2N_A)`
`7.5=(2xxm)/((500)^3xx10^(-30)xx6.02xx10^23)`
`M=(7.5xx(500)^3xx10(-30)xx6.02xx10^23)/2`
`M=282.18 " g/mol"`
`"Molar mass (M)"=("Mass of compound"xxN_A)/"Number of atoms"`
`282.18=(300xx6.02xx10^23)/"Number of atoms"`
Number of atoms = 6.4 x 1023
APPEARS IN
संबंधित प्रश्न
Distinguish between Hexagonal and monoclinic unit cells
Explain with reason sign conventions of ΔS in the following reaction
N2(g) + 3H2(g) → 2NH3(g)
An element has atomic mass 93 g mol−1 and density 11.5 g cm–3. If the edge length of its unit cell is 300 pm, identify the type of unit cell. (NA = 6.023 × 1023 mol−1)
Calculate the number of unit cells in 8.1 g of aluminium if it crystallizes in a f.c.c. structure. (Atomic mass of Al = 27 g mol–1)
What is the total number of atoms per unit cell in a face-centered cubic structure?
The number of atoms contained in a fcc unit cell of a monoatomic substance is ____________.
Edge length of unit cell of chromium metal is 287 pm with a bcc arrangement. The atomic radius is of the order:
An element with atomic mass 100 has a bcc structure and edge length 400 pm. The density of element is:
If a represents the edge length of the cubic systems, i.e. simple cubic, body centred cubic and face centered cubic, then the ratio of the radii of the sphere in these system will be:-
A solid is formed by 2 elements P and Q. The element Q forms cubic close packing and atoms of P occupy one-third of tetrahedral voids. The formula of the compound is ______.
