Solid State
Get insights from 127 questions on Solid State, answered by students, alumni, and experts. You may also ask and answer any question you like about Solid State
- 127
Questions
- 0
Discussions
- 0
Active Users
- 0
Followers
Have a question about Solid State? Ask our experts
Keep it short & simple. Type complete word. Avoid abusive language.
A compound is formed by two elements A and B. The element B forms cubic close packed structure and atoms of A occupy 1/3 of tetrahedral voids. If the formula of the compound is A?B?, then the value of x + y is in option
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
A compound is formed by two elements A and B. The element B forms cubic close packed structure and atoms of A occupy 1/3 of tetrahedral voids. If the formula of the compound is A?B?, then the value of x + y is in option
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
Since, atom B forms CCP structure. Therefore, there will be 4-B atoms.
Now, atom A occupies 1/3 of tetrahedral voids.
Hence, number of A atoms = 1/3 × 8 = 8/3
The correct formula of the compound = A? /? B?
= A? B?
x + y = 2 + 3 = 5
Choose the correction option for the total pressure (in atm.) in a mixture of 4g O2 and 2g H2 confined in a total volume of one litre at 0°C is:
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
Choose the correction option for the total pressure (in atm.) in a mixture of 4g O2 and 2g H2 confined in a total volume of one litre at 0°C is:
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
n (O? ) = 4/32 = 1/8 mol
n (H? ) = 2/2 = 1 mol
n (Total) = n (O? ) + n (H? ) = 1/8 + 1 = 9/8 mol
PV = nRT
P (Total) × 1 = (9/8) × 0.082 × 273
P (Total) = 25.18 atm
The molar conductance of NaCl, HCl and CH?COONa at infinite dilution are 126.45, 426.16 and 91.0 cm² mol¯¹ respectively. The molar conductance of CH?COOH at infinite dilution is. Choose the right option for your answer.
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
The molar conductance of NaCl, HCl and CH?COONa at infinite dilution are 126.45, 426.16 and 91.0 cm² mol¯¹ respectively. The molar conductance of CH?COOH at infinite dilution is. Choose the right option for your answer.
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
Λ°m (NaCl) = 126.45Scm² mol? ¹
Λ°m (HCl) = 426.16Scm² mol? ¹
Λ°m (CH? COONa) = 91Scm² mol? ¹
Λ°m (CH? COOH) =Λ°m (CH? COONa) +Λ°m (HCl)−Λ°m (NaCl)
= 91 + 426.16 – 126.45
= 391.72Scm² mol? ¹
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): In a particular point defect, an ionic solid is electrically neutral, even if few of its cations are missing from it unit cells.
Reason (R): In an ionic solid, Frenkel defect arises due to dislocation of cation from its lattice site to interstitial site, maintaining overall electrical neutrality.
In the light of the above statements, choose the most appropriate answer from the options given below:
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): In a particular point defect, an ionic solid is electrically neutral, even if few of its cations are missing from it unit cells.
Reason (R): In an ionic solid, Frenkel defect arises due to dislocation of cation from its lattice site to interstitial site, maintaining overall electrical neutrality.
In the light of the above statements, choose the most appropriate answer from the options given below:
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
Both statements are correct independently but the reason is not the correct reason of the assertion.
A certain element crystallises in a bcc lattice of unit cell edge length 27 Å . If the same element under the same conditions crystallises in the fcc lattice, the edge length of the unit cell in Å will be ___________. (Round off to the nearest Integer).
[Assume each lattice point has a single atom] [Assume √3 = 1.73, √2 = 1.41]
A certain element crystallises in a bcc lattice of unit cell edge length 27 Å . If the same element under the same conditions crystallises in the fcc lattice, the edge length of the unit cell in Å will be ___________. (Round off to the nearest Integer).
[Assume each lattice point has a single atom] [Assume √3 = 1.73, √2 = 1.41]
Edge length in bcc, a? = 27 Å
Let, Edge length in fcc be a? Å
Now, the same element crystallises in bcc as well as fcc.
For bcc: 4r = √3 a? ⇒ r = (√3 / 4) a?
For fcc: 4r = √2 a? ⇒ r = a? / (2√2)
So, (√3 / 4) a? = a? / (2√2)
(√3 / 4) × 27 = a? / (2√2)
a? = 33.13 Å
The nearest integer is 33.
Ga (atomic mass 70u) crystallizes in a hexagonal close packed structure. The total number of voids in 0.581 g of Ga is __________ ×10²¹. (Round off to the nearest integer).
[Given: N? = 6.023 ×10²³]
Ga (atomic mass 70u) crystallizes in a hexagonal close packed structure. The total number of voids in 0.581 g of Ga is __________ ×10²¹. (Round off to the nearest integer).
[Given: N? = 6.023 ×10²³]
First, the number of unit cells in the given mass is determined. In an HCP structure, there are 6 atoms and 18 total voids (6 octahedral + 12 tetrahedral) per unit cell. Multiplying the number of unit cells by 18 gives the total number of voids. The result is14.9 x 10²¹, which is rounded.
A hard substance melts at high temperature and is an insulator in both solid and in molten state. This solid is most likely to be a/ an:
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
A hard substance melts at high temperature and is an insulator in both solid and in molten state. This solid is most likely to be a/ an:
<!-- [if !supportLineBreakNewLine]-->
<!--[endif]-->
Covalent solids have high melting point due to strong covalent bonds, also they are insulator in both solid as well as in molten state.
In order to prepare a buffer solution of pH 5.74, sodium acetate is added to acetic acid. If the concentration of acetic acid in the buffer is 1.0 M, the concentration of sodium acetate in the buffer is _____ M. (Round off to the nearest integer).
[Given: pKa (acetic acid) = 4.74]
In order to prepare a buffer solution of pH 5.74, sodium acetate is added to acetic acid. If the concentration of acetic acid in the buffer is 1.0 M, the concentration of sodium acetate in the buffer is _____ M. (Round off to the nearest integer).
[Given: pKa (acetic acid) = 4.74]
For an acidic buffer solution, pH = pKa + log ( [Base]/ [Acid]).
Given pH = 5.74 and pKa = 4.74.
5.74 = 4.74 + log ( [Base]/1).
1 = log ( [Base]).
[Base] = 10M.
A 1 molal K4[Fe(CN)6] solution has a degree of dissociation of 0.4. Its boiling point is equal to that of another solution which contains 18.1 weight percent of a non electrolytic solute A. The molar mass of A is ______ u. (Round off to the nearest integer).
[Density of water = 1.0 g cm?³]
A 1 molal K4[Fe(CN)6] solution has a degree of dissociation of 0.4. Its boiling point is equal to that of another solution which contains 18.1 weight percent of a non electrolytic solute A. The molar mass of A is ______ u. (Round off to the nearest integer).
[Density of water = 1.0 g cm?³]
For the dissociation of K? [Fe (CN)? ]? 4K? + [Fe (CN)? ]? , the number of ions produced (n) is 5.
The degree of dissociation (α) is related to the van't Hoff factor (i) by α = (i-1)/ (n-1).
Given α = 0.4: 0.4 = (i - 1) / (5 - 1) => 1.6 = I - 1 => I = 2.6
Using the depression in freezing point formula

