Packing Fraction and Efficiency
Trending Questions
Q. 88.The ABAB close packing and ABC ABC close packing are respectively called as (1) hexagonal close packing(hcp) and cubic close packing (ccp) (2) ccp and hcp (3) body centred cubic(bcc) packing and hexagonal close packing (hcp) (4) hcp and bcc
Q. The density of Argon (fcc lattice) is 1.83g/cm³ at 20^° C. What is the length of an edge length
Q. 30 What is the coordination number in hcp lattice , fcc lattice, ccp lattice ?
Q. 4. Xenon crystallises in the face-centred cubic lattice and the edge length of the unit cell is 620 pm. The next nearest neighbour distance for the xenon is Options: 738.5 pm 620 pm 438.5 pm 310 pm
Q. 34. Total no. Of voids in 0.5 moles of a compound forming hcp structures are?
Q. An element crystallises in face-centred cubic lattice with density as 5 g/cm3 and edge length of the side of unit cell is 400 pm.
Take NA=6×1023 Then the molar mass of the element in g mol−1 is :
Take NA=6×1023 Then the molar mass of the element in g mol−1 is :
Q. A 5% (w/V) solution of cane sugar is isotonic with 0.878% (w/V) solution of urea. Calculate the molecular mass of urea if the molecular mass of cane sugar is 342 g mol−1.
- 60.05 g mol−1
- 76.1 g mol−1
- 45.5 g mol−1
- 120.1 g mol−1
Q. 7. B-is present in FCC lattice and A+ occupied octahedral void. If radius of B- is x . Calculate the maximum size of A+ that can fit octahedral void.
Q. For a NaCl unit cell, the Na+ occupies the edge centres and body centred positions of cube while Cl− occupies the corners and face centres of the cube.
Which is the correct expression for determining packing fraction (P.F.) of NaCl unit cell, if ions along an edge diagonal are absent?
Which is the correct expression for determining packing fraction (P.F.) of NaCl unit cell, if ions along an edge diagonal are absent?
Q. At 100∘C, copper (Cu) has fcc unit cell structure with cell edge length of x ∘A . what is the approximate density of Cu (in g cm−3) at this temperature? [Atomic mass of Cu =63.55 u]
(JEE Main 2019)
(JEE Main 2019)
- 205x3
- 105x3
- 211x3
- 422x3
Q. An element has a bcc structure having unit cells 12.08×1023. Find the total number of atoms in these unit cells.
- 36.24×1023
- 12.08×1023
- 6.02×1023
- 24.16×1023
Q. Coordination number in 3D hexagonal close packing is
Q. What percentage of crystal space in fcc arrangement is vacant?
- 74%
- 32%
- 68%
- 26%
Q. What is the coordination number of octahedral and tetrahedral voids in lattice ??
Q. how many nearest and next nearest neighbours respectively does potassium have in BCC lattice
Q. CsBr crystals are formed with bcc crystal lattice where Br− occupies the corners and Cs+ occupies the centre of the cube with a unit cell length of nearly 400 pm. What will be density of CsBr?
Given molecular mass of CsBr is 213 g/mol and
NA=6×1023
Given molecular mass of CsBr is 213 g/mol and
NA=6×1023
- 42.5 g/cm3
- 5.54 g/cm3
- 10.46 g/cm3
- 1.25 g/cm3
Q. A compound is formed by the elements X, Y . It has a cubic structure with atoms of element X occupying corners and half the face centres of the cube. Atoms of element Y are present at the remaining face centres and body centre.
What will be the simplest formula of the compound if all the atoms are removed from one body diagonal of the cube?
What will be the simplest formula of the compound if all the atoms are removed from one body diagonal of the cube?
Q. What percentage of crystal space in fcc arrangement is vacant?
- 74%
- 32%
- 68%
- 26%
Q.
Fraction of total volume occupied by atoms in a simple cubic cell is
Q. Calculation of packing efficiency in hcp(hexagonal closed packing) mathematically.
Q. Are all octahedral voids in a HCP lattice inside the lattics
Q.
Let's take a Rubik's cube.
We know that a Rubik's cube is made up of 27 identical smaller cubes which make up one large cube as seen in the picture above.
Let us assume that the one of the small cubes is missing. What will be the packing fraction in this case?
127
0
1
2627
Q. Packing fraction is maximum in .
- simple cubic
- face centred cubic
- body centred cubic
Q. If a 100 mL of urea solution is isotonic with 10% (w/V) solution of cane sugar , then calculate the weight of urea in the urea solution
Given: The molecular mass of cane sugar is 342 g mol−1 and that of urea is 60 g mol−1.
Given: The molecular mass of cane sugar is 342 g mol−1 and that of urea is 60 g mol−1.
- 1.01 g
- 0.54 g
- 2.58 g
- 1.75 g
Q. 1.82 g of a metal requires of 1 N HCL to dissolve it. What volume is equivalent weight of metal??
Q.
What you mean by packing fraction in 3D?
Space in the unit cell
Volume of unit cell
Space occupied by particles/ Total volume of cell.
Density of particles/Density of cell.