A) 15.4 kg/mol
B) \[1.54\times {{10}^{4}}\,kg/mol\]
C) \[3.08\times {{10}^{4}}\,kg/mol\]
D) \[3.08\times {{10}^{3}}kg\,/mol\]
Correct Answer: A
Solution :
Specific volume (volume of 1 g) cylindrical virus particle = \[6.02\times {{10}^{-2}}\,cc/g\] Radius of virus \[(r)=7\times {{10}^{-8}}\,cm\] Length of virus = \[10\times {{10}^{-8}}\,cm\] Volume of virus \[=\pi {{r}^{2}}\ell =\frac{22}{7}\times {{(7\times {{10}^{-8}})}^{2}}\times 10\times {{10}^{-8}}\] \[=154\,\times {{10}^{-23}}cc\] Weight of one virus particle \[=\frac{volume}{specific\,volume}=\frac{154\times {{10}^{-23}}}{6.02\times {{10}^{-2}}}\] \[\therefore \,Mol.\,wt.\,of\,virus=Wt.\,of\,{{N}_{A}}\,particle\] \[=\frac{154\times {{10}^{-23}}}{6.02\times {{10}^{-2}}}\times 6.02\times {{10}^{23}}\] \[=15400\,g/mol=15.4\,kg/mol\]
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