An element M crystallises in a body centred cubic unit cell with a cell edge of 300 pm. The density of the element is 6.0 g cm–3. The number of atoms present in 180 g of the element is ______× 1023. (Nearest integer)
Solution and Explanation
The correct answer is 22
M is body centred cubic, ∴ Z=2
Let mass of 1 atom of M is A
Edge length =300pm
Density =\(6 g/cm^3\)
∴ \(6 g/cm^3=\frac{Z×A}{300×10^{-10^3}}=\frac{2×A}{27×10^{-24}}\)
\(A=81×10^{-24} g\)
Atoms of M=\(\frac{Total \space mass}{Mass\space of \space one \space atom}\)
= \(\frac{180}{81×10^{-24}}\)
= \(22.22×10^{23}\)
≅ \(22×10^{23}\)
Top Questions on The solid state
- Which of the following crystals has the unit cell such that \( a = b \neq c \) and \( \alpha = \beta = 90^\circ, \gamma = 120^\circ \)?
- KCET - 2024
- Chemistry
- The solid state
- MnO exhibits:
- KCET - 2024
- Chemistry
- The solid state
- The number of atoms in 4.5 g of a face-centred cubic crystal with edge length 300 pm is:
- KCET - 2024
- Chemistry
- The solid state
- Which of the following material is not a semiconductor.
- JEE Main - 2024
- Chemistry
- The solid state
- In a metal deficient oxide sample, MXY2O4(M and Y are metals), M is present in both +2 and +3 oxidation states and Y is in +3 oxidation state. If the fraction of M2+ ions present in M is \(\frac{1}{3}\), the value of X is______.
- JEE Advanced - 2024
- Chemistry
- The solid state
Questions Asked in JEE Main exam
- A rectangular loop of length 2.5 m and width 2 m is placed at 60° to a magnetic field of 4 T. The loop is removed from the field in 10 sec. The average emf induced in the loop during this time is:
- JEE Main - 2024
- Electromagnetic induction
- Two coils have mutual inductance 0.002 H. The current changes in the first coil according to the relation \( i = i_0 \sin \omega t \), where \( i_0 = 5 \, \text{A} \) and \( \omega = 50\pi \, \text{rad/s} \). The maximum value of emf in the second coil is \( \frac{\pi}{\alpha} \). The value of \( \alpha \) is ______.
- JEE Main - 2024
- Electromagnetic induction
- The 20th term from the end of the progression \[ 20, 19, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \ldots, -129 \frac{1}{4} \] is:
- JEE Main - 2024
- Arithmetic Progression
- We have three aqueous solutions of NaCl labelled as ‘A’, ‘B’ and ‘C’ with concentration 0.1 M, 0.01M & 0.001 M, respectively. The value of vant’ Haft factor (i) for these solutions will be in the order.
- JEE Main - 2024
- Colligative Properties
- $\Delta_{vap}H^\circ$ for water is $+40.49 \, \text{kJ mol}^{-1}$ at 1 bar and $100^\circ\text{C}$. Change in internal energy for this vaporisation under same condition is .............. kJ mol$^{-1}$. (Integer answer)
(Given R = 8.3 J K$^{-1}$ mol$^{-1}$)- JEE Main - 2024
- Enthalpy change
Concepts Used:
Solid State
Solids are substances that are featured by a definite shape, volume, and high density. In the solid-state, the composed particles are arranged in several manners. Solid-state, in simple terms, means "no moving parts." Thus solid-state electronic devices are the ones inclusive of solid components that don’t change their position. Solid is a state of matter where the composed particles are arranged close to each other. The composed particles can be either atoms, molecules, or ions.
Types of Solids:
Based on the nature of the order that is present in the arrangement of their constituent particles solids can be divided into two types;
- Amorphous solids behave the same as super cool liquids due to the arrangement of constituent particles in short-range order. They are isotropic and have a broad melting point (range is about greater than 5°C).
- Crystalline solids have a fixed shape and the constituent particles are arranged in a long-range order.