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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. (a)`3.08xx10^(3) kg//mol`B. (b)`15.4 kg//mol`C. (c )`1.54xx10^(4) kg//mol`D. (d)`3.08xx10^(4) kg//mol` |
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Answer» Correct Answer - B Specific volume `("volume of 1 g")` of cylindrical virus particle `=6.02xx10^(-2)` "c c/g" Radius of virus (r )`=7Å=7xx10^(-8) cm` Length of virus`=10xx10^(-8)cm` Volume of virus `pir^(2)=22/7xx(7xx10^(-8))^(2)xx10xx10^(-8)=154xx10^(-23)` c c Wt. of one virus particle`=("volume")/("specific volume")` `:.` Mol. wt. of virus=wt. of `N_(A)` particle `=(154xx10^(-23))/(6.02xx10^(-2))xx6.02xx10^(23)` `=15400 g//mol=15.4 kg//mol` |
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