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