Solid State

The coordination number of an atom in a body- centred cubic structure is___________.

[Assume that the lattice is made up of atoms]

0 Follower 1 View

Contributor-Level 10

Coordination no. of an atom in BCC is 8.

0 0

New answer posted

2 months ago

The number of octahedral voids per lattice site in a lattice is………. (Rounded off to the nearest integer)

0 Follower 3 Views

Contributor-Level 10

Effective no. of octahedral voids in a lattice = n

Effective no. of lattice point in a lattice = n

$∴$ Ratio = n/n = 1

0 0

New answer posted

2 months ago

The solubility product of Pbl₂ is 8.0 * 10^-⁹. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x * 10^-⁶ mol/L. The value of x is…………………… (Rounded off to the nearest integer)

[Given $\sqrt[]{2} = 1.41]$

0 Follower 5 Views

Contributor-Level 10

$P b l_{2} ? P b^{2 +} + 2 l^{-} K_{s q} = 8.0 * 1 0^{- 9}$

Pb (NO₃)₂ -> Pb²⁺ +^{$2 N O_{3}^{-}$}

0.1 M-

-0.1 M 0.1 M

$K_{s q} = 8 * 1 0^{- 9} U s i n g K_{s q} = [P b^{2 +}] {[I^{-}]}^{2}$

$8 * 1 0^{- 9} = 0.1 * {(2 S)}^{2}$

S = 141 * 10^-6 M

0 0

New answer posted

2 months ago

The unit cell of copper corresponds to a face centered cube of edge length 3.596 $\overset{o}{A}$ with one copper atom at each lattice point. The calculated density of copper in kg/m³ is…….

[Molar mass of Cu = 63.54g : Avogadro Number = 6.022 * 10²³]

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

Density = $\frac{M * z}{N_{A} a^{3}}$

$d = \frac{63.5 * 4}{6.02 * 1 0^{23} * {(3.596 * 1 0^{- 10})}^{3} * 1000} = 9076 k g / m^{3}$

0 0

New answer posted

3 months ago

An element M crystallises in a body centred cubic unit cell with a cell edge of 300 pm. The density of the element is 6.0g cm^-³. The number of atoms in 180g of the element is_________* 10²³ (Nearest integer)

0 Follower 2 Views

Contributor-Level 10

B.C.C structure

a = 300 pm = 300 * 10^-¹² m

d = 6g/cm³

z = 2

$d = \frac{Z * M}{a^{3}}$

$6 = \frac{2 * A}{{(300 * 1 0^{- 10})}^{3}} = \frac{2 * A}{27 * 1 0^{- 24}}$

$∴ A t o m s o f M$ = 3.69 * 6.022 * 10²³

= 22.22 * 10²³

the nearest integer = 22

0 0

New answer posted

3 months ago

A copper complex crystallizing in a CCP lattice with a cell edge of 0.4518 nm has been revelaed by employing X-ray diffraction studies. The density of a copper complex is found to be 7.26 g cm^-3. The molar mass of copper complex is___________ g mol^-1. (Nearest integer)

[Given : N_A = 6.022 * 10²³ mol^-1]

0 Follower 2 Views

Contributor-Level 10

Density, $ρ = 7.62 g c m^{- 3}$

Cell edge, a = 0.4518nm = 0.458 * 10^-7 cm

Effective number of atoms, Z = 4 for CCP.

Using ; $ρ = \frac{Z * A}{N_{A} * a^{3}}$

$ρ = \frac{4 * A}{6.022 * 1 0^{23} * {(0.458 * 1 0^{- 7})}^{3}}$

$7.62 = \frac{4 * A}{6.022 * {(0.4518)}^{3} * 1 0^{2}}$

A = 105.79 g/mol

Ans. is 106 (the nearest integers)

0 0

New answer posted

3 months ago

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: Sharp glass edge becomes smooth on heating it upto its melting point.

Reason R: The viscosity of glass decreases on melting.

Choose the most appropriate answer from the options given below.

0 Follower 1 View

Contributor-Level 10

Both statement are correct for the glass body heating, but reason is not correct explanation during heating process of glass, constituents unit rupture of glass body and gives the edge smoothness.

0 0

New answer posted

3 months ago

Diamond has a three dimensional structure of C atoms formed by covalent bonds. The structure of diamond has face centre cubic lattice where 50% of the tetrahedral voids are also occupied by carbon atoms. The number of carbon atoms present per unit cell of diamond is_________.

0 Follower 1 View

Contributor-Level 10

In fcc structure of diamond four C present in fcc lattice and other four C present in tetrahedral voids where 50% of tetrahedral voids are occupied. Hence number of carbon atoms present per unit cell of diamond is 8.

0 0

New answer posted

3 months ago

A 50 watt bulb emits monochromatic red light of wavelength of 795 nm. The number of photons emitted per second by the bulb is x * 10²⁰. The value of x is________.

[Given: h = 6.63 * 10^-34 Js and c = 3.0 * 10⁸ ms^-1]

0 Follower 2 Views

Contributor-Level 10

Power = 50 watt = 50 J/sec

Energy emitted per second = 50 J

Wavelength ; $λ$ = 795 nm

Energy of one photon =

$= 0.025 * 1 0^{- 17} J$

Number of photons emitted per second = $\frac{50}{0.025 * 1 0^{- 17}}$

= 2 * 10²⁰

0 0

New answer posted

3 months ago

The empirical formula for a compound with a cubic close packed arrangement of anions and with cations occupying all the octahedral sites is A_xB. The value of x is _______.

0 Follower 2 Views