Silver crystallises in fcc lattice- If edge length of the cell is 4-07 - 10-8 cm and density is 10-5 g cm -3- calculate the atomic mass of silver-

It is given that the edge length, a = 4.077 × 10−8 cm Density, d = 10.5 g cm−3 As the lattice is fcc type, the number of atoms per unit cell, z = 4 We also kno ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Chemistry

Why Classification?

Silver crysta...

Question

Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10⁻⁸ cm and density is 10.5 g cm⁻³, calculate the atomic mass of silver.

Open in App

Solution

It is given that the edge length, a = 4.077 × 10⁻⁸ cm

Density, d = 10.5 g cm⁻³

As the lattice is fcc type, the number of atoms per unit cell, z = 4

We also know that, N_A = 6.022 × 10²³ mol⁻¹

Using the relation:

= 107.13 gmol⁻¹

Therefore, atomic mass of silver = 107.13 u

Suggest Corrections

Similar questions

Q. Silver crystallises in face-centred cubic lattice. If edge length of the unit cell is

4.07 \times 10^{- 8} c m

and density of silver is

10.48 g c m^{- 3}

, determine the relative atomic mass of silver.

Q. Silver crystallises in fcc lattice. If edge length of the cell is

4.077 \times 10^{8}

cm and density is

10.5

g cm

^{- 3}

, calculate the atomic mass (g) of silver.

Q. Silver

(A g)

crystelise in

F C C

lathice if edge length

(a)

of the cell is

4.07 \times 10^{- 8} c m

& density is

10.5 g / {c m}^{3}

. Calculate the atomic mass of

A g

Q. Silver crystallises in a face-centred cubic in cell. The density of

A g

10.5 g c m^{- 3}

. Calculate the edge length of the unit cell.

Q. Silver crystallises in FCC structure. If density of silver is

10.51 g . {c m}^{- 3}

. Calculate the volume of unit cell. [Atomic mass of silver

(A g) = 108 g m^{- 1}

]

Join BYJU'S Learning Program

Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10−8 cm and density is 10.5 g cm−3, calculate the atomic mass of silver.

It is given that the edge length, a = 4.077 × 10−8 cm Density, d = 10.5 g cm−3 As the lattice is fcc type, the number of atoms per unit cell, z = 4 We also know that, NA = 6.022 × 1023 mol−1 Using the relation: = 107.13 gmol−1 Therefore, atomic mass of silver = 107.13 u

Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10⁻⁸ cm and density is 10.5 g cm⁻³, calculate the atomic mass of silver.

It is given that the edge length, a = 4.077 × 10⁻⁸ cm

Density, d = 10.5 g cm⁻³

As the lattice is fcc type, the number of atoms per unit cell, z = 4

We also know that, N_A = 6.022 × 10²³ mol⁻¹

Using the relation:

= 107.13 gmol⁻¹

Therefore, atomic mass of silver = 107.13 u