Solid State Question 264
Question: The density of solid argon (Ar = 40 g/mol) is 1.68 g/mL at 40 K. If the argon atom is assumed to be a sphere of radius $ 1.50\times {{10}^{-8}} $ cm, what % of solid Ar is apparently empty space- (use $ N _{A} $ = $ 6\times 10^{23} $ )
Options:
A) 35.64
B) 64.36
C) 74
D) none of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ 1cm^{3}=1ml $
$ \Rightarrow d=1.68g/cm^{3}, $ mol $ wt=40, $
$ N _{A}=6\times 10^{23} $
$ \Rightarrow \frac{d\times N _{A}}{mol.wt}=\frac{Z}{a^{3}} $ Efficiency $ =\frac{Z\times \frac{4}{3}\pi R^{3}}{a^{3}}=\frac{d\times N _{A}\times \frac{4\pi }{3}\pi R^{3}}{mol.wt} $
$ =\frac{1.68\times 6\times 10^{23}\times 4\times 3.14\times {{(1.50\times {{10}^{-8}})}^{3}}}{40\times 3} $
$ =0.35604 $ % efficiency = 35.604%
$ \Rightarrow void%=100-35.604%=64.39% $
BETA
