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Specific volume of cylindrical virus particle is `6.02xx10^(-2) c c//g` whose radius and length `7 Å` and `10 Å` respectively. If `N_(A)=6.02xx10^(23)`, find molecular weight of virus:A. `15.4 kg//mol`B. `1.54xx10^(4)kg//mol`C. `3.08xx10^(4)kg//mol`D. `3.08xx10^(3)kg//mol` |
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Answer» Correct Answer - A Specific volume (volume of 1 g )cylindrical virus particle `=6.02xx10^(-2) c c//g` Radius of virus ( r )`=7xx10^(-8)cm` Length of virus (1) `=10xx10^(-8)cm` Volume of virus `=pir^(2)1=(22)/(7)xx(7xx10^(-8))^(2)xx10xx10^(-8)` `=154xx10^(-23)c c` weight of one virus particle `=("Volume")/("Specific volume")=(154xx10^(-23))/(6.02xx10^(-2))` `:.` Molecular weight of virus = weight of `N_(A)` particles `=(154xx10^(-23))/(6.02xx10^(-2))xx6.023xx10^(23)` `=15400g//mol=15.4 kg//mol` |
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