Solid State - Result Question 7
7. A solid having density of $9 \times 10^{3} $ $kg$ $ m^{-3}$ forms face centred cubic crystals of edge length $200 \sqrt{2}$ $ pm$. What is the molar mass of the solid?
[Avogadro constant $\left.=6 \times 10^{23} mol^{-1}, \pi=3\right]$
(2019 Main, 11 Jan I)
(a) $0.03050 $ $kg$ $ mol^{-1}$
(b) $0.4320 $ $kg$ $ mol^{-1}$
(c) $0.0432 $ $kg$ $ mol^{-1}$
(d) $0.0216 $ $kg$ $ mol^{-1}$
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Answer:
Correct Answer: 7. (a)
Solution:
- Density of a crystal
$ d=\dfrac{M \times Z}{N _A \times a^{3}} \Rightarrow M=\dfrac{d \times N _A \times a^{3}}{Z} $
Given, $d=9 \times 10^{3} $ $kg$ $ m^{-3}$
$ \begin{aligned} M & =\text { Molar mass of the solid } \\ Z & =4(\text { for fcc crystal }) \\ N _A & =\text { Avogadro’s constant }=6 \times 10^{23}\hspace{2mm} mol^{-1} \\ a & =\text { Edge length of the unit cell } \\ & =200 \sqrt{2} \hspace{2mm} pm=200 \sqrt{2} \times 10^{-12} m \end{aligned} $
On substituting all the given values, we get
$ \begin{aligned} & =\dfrac{\left(9 \times 10^{3}\right) kg \hspace{2mm} m^{-3} \times\left(6 \times 10^{23}\right) mol^{-1} \times\left(200 \sqrt{2} \times 10^{-12}\right)^{3} m^{3}}{4} \\ & =0.0305 \hspace{2mm} kg \hspace{2mm} mol^{-1} \end{aligned} $
BETA
