Saved Bookmarks
| 1. |
A plant virus was found to consist of uniform cylindrical particles `100A` in diameter and `4000 A` long The virus has a specific volume `0.314 cm^(3) g^(-1)` If the virus particle is considered to be one molecule, what is its molecular weight ? . |
|
Answer» `V = pi r^(2) h = (3.14) ((100Å)/(2))^(2) ( 400Å)` ` = 3.14 xx 10^(7) (Å)^(3) ((10^(-8)cm)/(Å))^(3)` `= 3.14 xx 10^(-17) cm^(3)` Therefore, the specific volume is `0.314 cm^(3) g^(-1)` of molecule. If the specific volume is `3.14 xx 10^(-17)cm^(3)` The weight per molecule is `=((1g)/(0.314cm^(3)))(3.14 xx 10^(-7) cm^(3))` ` = 10^(-16) "molecule"^(-1)` Molecular weight of virus `=(10^(-6) "g molecule"^(-1))` `((6.02 xx 10^(23 "molecules"))/(mo1))` ` =6.02 xx 10^(7) g mo1^(-1)` . |
|