NTA Abhyas 2020
- Exam
- NTA Abhyas
- Year
- 2020
- Subject
- Chemistry
- Chapter
- Solid State
NTA Abhyas 2020 · Chemistry · Solid State
Calcium crystallizes in a face centred cubic unit cell with a...
Calcium crystallizes in a face centred cubic unit cell with a & The density of the metal if it contains 0.1% schottky defects would be:
Options
- A\text{1}{\text{.51 gcm}^{\text{-3}}
- B\text{2}{\text{.51 gcm}^{\text{-3}}
- C\text{15}{\text{.1 gcm}^{\text{-3}}
- D\text{0}{\text{.151 gcm}^{\text{-3}}
Correct answer
A. \text{1}{\text{.51 gcm}^{\text{-3}}
Step-by-step solution
A face centred cubic unit cell contains four atoms. Schottky defect is basically a vacancy defect in ionic solids. Due to Schottky defects, the vacant spaces will increase, resulting in decrease in number of atoms per unit cell.
Thus, number of atoms will reduce to:
The density of the metal can be calculated by:
Where, is the molecular mass of calcium {\text{= 40 g mol}^{\text{-1}}
is the Avogadro number = 6.022 \times {10}^{23} {\text{mol}^{-} {}^{1}
is the edge length \text{= 0}\text{.560 nm}\text{=}\text{0}\text{.56}\text{\times}{\text{10}^{- \text{7}} \text{cm}
Putting the values in the formula above:
\text{=}\dfrac{\text{3}\text{.996}\text{\times}\text{(40}\text{g}{\text{mol}^{- \text{1}} \text{)}}{{\text{(0}\text{.560}\text{\times}{\text{10}^{- \text{7}} \text{cm)}^{3} \text{\times}\text{(6}\text{.022}\text{\times}{\text{10}^{\text{23}} {\text{mol}^{- \text{1}} \text{)}}
= 1.51\text{g}{\text{cm}^{- 3}
Thus, number of atoms will reduce to:
The density of the metal can be calculated by:
Where, is the molecular mass of calcium {\text{= 40 g mol}^{\text{-1}}
is the Avogadro number = 6.022 \times {10}^{23} {\text{mol}^{-} {}^{1}
is the edge length \text{= 0}\text{.560 nm}\text{=}\text{0}\text{.56}\text{\times}{\text{10}^{- \text{7}} \text{cm}
Putting the values in the formula above:
\text{=}\dfrac{\text{3}\text{.996}\text{\times}\text{(40}\text{g}{\text{mol}^{- \text{1}} \text{)}}{{\text{(0}\text{.560}\text{\times}{\text{10}^{- \text{7}} \text{cm)}^{3} \text{\times}\text{(6}\text{.022}\text{\times}{\text{10}^{\text{23}} {\text{mol}^{- \text{1}} \text{)}}
= 1.51\text{g}{\text{cm}^{- 3}