A) 32%
B) 54%
C) 68%
D) 62%
Correct Answer: D
Solution :
Volume of one atom \[=\frac{4}{3}\pi {{r}^{3}}\] \[=\frac{4}{3}\times \frac{22}{7}\times {{(1.54\times {{10}^{-8}})}^{3}}c{{m}^{3}}\] \[=1.53\times {{10}^{-23}}c{{m}^{3}}\] Volume of all atoms in 1.65 g Ar \[=\frac{165}{40}\times 6.02\times {{10}^{23}}\times 1.53\times {{10}^{-23}}c{{m}^{3}}\] \[=0.380\text{ }c{{m}^{3}}\] Volume of solid Ar containing \[1.65\text{ }g=1\text{ }c{{m}^{3}}\] Empty space\[=1-0.380=0.620\] % = 62%
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