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