Density
Trending Questions
(Given: pKw=13, pKa[HA]=5, NA=6×1023)
- 43%
- 30%
- 25%
- 45%
- Copper (atomic mass= 63 g/mol)
- Silver (atomic mass = 108 g/mol)
- Molybdenum (atomic mass = 95 g/mol)
- Gold (atomic mass = 196 g/mol)
- 314 pm
- 136 pm
- 152 pm
- 111 pm
- both (A) and (B)
- applying high pressure
- increase in temperature
- no effect on coordination
(Round off x to the nearest integer).
[Given=NA=6.023×1023]
(JEE Main 2021)
A metal crystallises into two cubic phases, fcc and bcc. These have unit cell lengths 3.5 and 3 angstrom respectively. Calculate the ratio of densities of fcc and bcc.
4.56
2.76
1.26
2.87
A liquid X of specific heat capacity and at is mixed with a liquid Y of specific heat capacity and at , when the final temperature recorded is . Find in what proportion the weights of the liquids are mixed.
- Iron
- Copper
- Gold
- Silver
State the condition under which Hydrogen reacts with Nitrogen.
(use NA=6×1023):
- 4Fe2+ and 4O2−
- 2Fe2+ and 2O2−
- 1Fe2+ and 1O2−
- 3Fe2+ and 4O2−
Given: [At. wt. of Fe = 56 g mol−1]
The metal calcium crystallizes in F.C.C unit cell with a = 600 pm. The density of metal if it contains 0.2% schottky deficits.
1.22 g/cc
5 g/cc
20 g/cc
50 g/cc
- 1.069
- 0.917
- 0.725
- 1.231
- 1.73×1026
- 1.22×1026
- 1.38×1026
- 1.73×1025
- 2
- 3
- 4
- 5
(use NA=6×1023):
- 4Fe2+ and 4O2−
- 2Fe2+ and 2O2−
- 1Fe2+ and 1O2−
- 3Fe2+ and 4O2−
- 1.73×1026
- 1.22×1026
- 1.38×1026
- 1.73×1025
- Oxygen
- Ammonia
- Nitrogen
- Nitrous oxide
A metal crystallises in two cubic phases, face entered cubic and body entered cubic whose unit length are 3.5 and 3.0 Angstrom respectively. Calculate the ratio of density of FCC and BCC.
1.459
1.259
1.159
1.359
a=6∘A, b=4∘A and c=7 ∘A.
If the molar mass is 21 g, calculate the density of crystal.
- 0.99 g/cm3
- 0.83 g/cm3
- 1.25 g/cm3
- 1.50 g/cm3
- 4×1025
- 3×1025
- 2×1025
- 1×1025
The metal calcium crystallizes in F.C.C unit cell with a = 600 pm. The density of metal if it contains 0.2% schottky deficits.
1.22 g/cc
5 g/cc
20 g/cc
50 g/cc
(Take NA=6×1023)
- 2
- 3
- 4
- 5
- (NH4)2SO4
- NH4NO2
- (NH4)2(CO3)
- (NH4)2Cr2O7
- Iron
- Copper
- Gold
- Silver
Now let's calculate the mass of a 3-D unit cell. Assume that each sphere has a mass m and the sphere lies at the center of the faces of the cube. What will be the mass in the following case?
2m
4m
m
12m
- SC
- BCC
- FCC
- HCP
[Atomic mass of Cu =63.55 u]