HSC Science (Computer Science) 12th Board ExamMaharashtra State Board
Share
Notifications

View all notifications

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. - HSC Science (Computer Science) 12th Board Exam - Chemistry

Login
Create free account


      Forgot password?

Question

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.

Solution

Edge length of unit cell (a) = 288pm

volume of unit cell = (a)3

=(288)3 = 2.389 × 10-23 cm3

density of iron = ρ = 7.86g.cm3

Mass of iron unit cell (M) = Density × Volume

= 7.86 × 2.389 × 10-23

= 18.778 × 10-23 g

molar mass of iron = 56 g.mol-1

moles of iron in a unit cell = mass of iron in unit cell / molar mass of iron

= 18.778 × 10-23 / 56

= 3.353 × 10-24 mol

number of atoms per unit cell = moles of iron in a unit cell × Avogadro's number

= 3.353 × 10-24 × 6.022 x 1023

= 2.0191

≈ 2 atoms per unit cell

In a body centered cubic structure (bcc), the total number of atoms equals 2.

  Is there an error in this question or solution?

APPEARS IN

 2013-2014 (October) (with solutions)
Question 4.1 | 7.00 marks

Video TutorialsVIEW ALL [1]

Solution 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. Concept: Number of Atoms in a Unit Cell.
S
View in app×