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