A) 5.584 \[\overset{\text{o}}{\mathop{\text{A}}}\,\]
B) 4.285 \[\overset{\text{o}}{\mathop{\text{A}}}\,\]
C) 47.392 \[\overset{\text{o}}{\mathop{\text{A}}}\,\]
D) 1.121 \[\overset{\text{o}}{\mathop{\text{A}}}\,\]
Correct Answer: B
Solution :
[b]| Molar mass of \[CsBr\] |
| \[=212.8g\,mo{{l}^{-\,1}}\] |
| \[\therefore \]\[\text{Molar}\,\text{volume=}\frac{\text{molar}\,\text{mass}}{\text{density}}\] |
| \[=\frac{212.8}{4.49}=47.39\,c{{m}^{3}}\] |
| \[6.023\times {{10}^{23}}\]molecules are there in \[47.39\,c{{m}^{3}}\] |
| 1 molecule is there in\[\frac{47.39}{6.023\times {{10}^{23}}}\] |
| \[=7.868\times {{10}^{-\,23}}c{{m}^{3}}\] |
| \[\therefore \]Volume of the unit cell |
| \[={{a}^{3}}=7.868\times {{10}^{-\,23}}c{{m}^{3}}\] |
| Side of the unit cell |
| \[=a={{(7.868\times {{10}^{-\,23}})}^{1/3}}\] |
| \[=4.285\times {{10}^{-\,8}}cm\] |
| \[=4.285\overset{{}^\circ }{\mathop{A}}\,\] |
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