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 |
\[\text{=}\frac{\text{volume}}{\text{specific}\,\text{volume}}\text{=}\frac{154\times {{10}^{-23}}}{6.02\times {{10}^{-2}}}\] |
\[\therefore \] 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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