| Specific volume of cylindrical virus particle is \[6.02\times {{10}^{-2}}cc/g.\] whose radius and length are 7 \[\overset{\text{o}}{\mathop{\text{A}}}\,\] & 10 \[\overset{\text{o}}{\mathop{\text{A}}}\,\] respectively. |
| If \[{{N}_{A}}=6.02\times {{10}^{23}}mo{{l}^{-1}}\] find molecular weight of virus |
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) \[15.4kg/mol\]
Correct Answer: D
Solution :
| [d] Specific volume (volume of 1 g) of cylindrical virus particle \[=6.02\times {{10}^{-2}}cc/g\] |
| Radius of virus \[=7\overset{\text{o}}{\mathop{\text{A}}}\,=7\times {{10}^{-8}}cm\] |
| Length of virus \[=10\times {{10}^{-8}}cm\] |
| Volume of virus = \[\pi {{r}^{2}}I=\frac{22}{7}\times {{(7\times {{10}^{-~8}})}^{2}}\times 10\times {{10}^{-}}^{8}\]\[=154\times {{10}^{-23}}cc\] |
| \[Wt.\text{ }of\text{ }one\text{ }virus\text{ }particle=\frac{volume}{specific\text{ }volume}\] |
| \[\therefore \]Mol. wt. of virus = Wt. of \[{{N}_{A}}\]particle |
| \[=\frac{154\times {{10}^{-23}}}{6.02\times {{10}^{-2}}}\times 6.02\times {{10}^{23}}=15400g/mol\] |
| \[=15.4kg/mol\] |
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