मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

The Density of Silver Having an Atomic Mass of 107.8 G Mol- 1 is 10.8 G Cm-3. If the Edge Length of Cubic Unit Cell is 4.05 × 10- 8 Cm, Find the Number of Silver Atoms in the Unit Cell.

Advertisements
Advertisements

प्रश्न

The density of silver having an atomic mass of 107.8 g mol- 1 is 10.8 g cm-3. If the edge length of cubic unit cell is 4.05 × 10- 8
 cm, find the number of silver atoms in the unit cell.
 ( NA = 6.022 × 1023, 1 Å = 10-8 cm)

संख्यात्मक
Advertisements

उत्तर

Given:
Density (d) = 10.8 g cm-3
Edge length (a) = 4.05 x 10- 8 cm
Molar mass = 107.8 g mol-1
Avogadro's number (NA) = 6.022 x 1023

To find:
Number of atoms in the unit cell

Formula:
a. Mass of one atom = `"Atomic mass"/"Avogadro number"`
b. Volume of unit cell = a3
c. Density = `"Mass of unit cell"/"Volume of unit cell"`

Calculation:
a) Mass of one Ag atom = `"Atomic mass of Ag"/"Avogadro number"`
Avogadro number
= `107.8/(6.022 xx 10^23)`

= 1.79 x 10-22 g

b) Volume of unit cell = a3
= ( 4.05 x 10-8 )3
= 6.64 x 10-23 cm3

c)
Density (d) = `"Mass of unit cell"/"Volume of unit cell"`

= `"Number of atoms in unit cell x Mass of one atom"/"Volume of unit cell"`

10.8 = `("Number of atoms in unit cell" xx 1.79 xx 10^-22)/(6.64 xx 10^-23)`

Number of atoms in unit cell = `( 10.8 xx 6.64 xx 10^-23)/( 1.79 xx 10^-22)`

= 40.06 x 10-1 = 4.0 ≈ 4

∴ The number of atoms in the unit cell of silver is 4.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2017-2018 (July) Set 1

APPEARS IN

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

A unit cell of iron crystal has edge length 288 pm and density 7.86 g.cm-3. Find the number of atoms per unit cell and type of the crystal lattice.

Given : Molar mass of iron = 56 g.mol-1; Avogadro's number NA = 6.022 x 1023.


Distinguish between Hexagonal and monoclinic unit cells


Distinguish between Face-centred and end-centred unit cells.


An element with molar mass 2.7 × 10-2 kg mol-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 × 103 kg m−3, what is the nature of the cubic unit cell?


Explain with reason sign conventions of ΔS in the following reaction

N2(g) + 3H2(g) → 2NH3(g)


A face centred cube (FCC) consists of how many atoms? Explain


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)


An element forms a cubic unit cell with edge length 405 pm. Molar mass of this element is 2.7 × 10−2 kg/mol and its density is given as 2.7 × 103 kg/m3. How many atoms of these elements are present per unit cell?


A substance forms face-centered cubic crystals. Its density is 1.984 g/cm3 and the length of the edge of the unit cell is 630 pm. Calculate the molar mass in g/mol?


A metal has a body-centered cubic crystal structure. The density of the metal is 5.96 g/cm3. Find the volume of the unit cell if the atomic mass of metal is 50.


An element (atomic mass 100 g/mol) having bcc structure has unit cell edge 400 pm. The density of element is (No. of atoms in bcc, Z = 2).


Sodium metal crystallises in a body-centred cubic lattice with a unit cell edge of 4.29 Å. The radius of the sodium atom is approximately ______.


The empty space in the body-centered cubic lattice is ____________.


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:


The density of a metal which crystallises in bcc lattice with unit cell edge length 300 pm and molar mass 50 g mol−1 will be:


An element with atomic mass 100 has a bcc structure and edge length 400 pm. The density of element is:


The percentage of empty space in a body centred cubic arrangement is ______.


The correct set of quantum numbers for 2p sub shell is:


Percentage of free space in body centred cubic unit cell is


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 ______.


In an ionic solid r(+) = 1.6 Å and r(−) = 1.864 Å. Use the radius ratio rule to the edge length of the cubic unit cell is ______ Å.


The ratio of number of atoms present in a simple cubic, body-centred cubic and face-centred cubic structure are, respectively ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×