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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