A) 35.64
B) 64.36
C) 74
D) none of these
Correct Answer: B
Solution :
[b] \[1c{{m}^{3}}=1ml\] \[\Rightarrow d=1.68g/c{{m}^{3}},\]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% \[\Rightarrow void%=100-35.604%=64.39%\]
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