Question

# Specific volume of cylindrical virus particle is 6.02×10−2ccg, whose radius and length arc 7∘A respectively. If NA=6.02×1023, find molecular weight of virus.

Solution

## Specific volume =6.02×10−2ccgm So density =1specific volume=16.02×10−2gmcc Volume of the cylinder =πr2∗h=π×7×7×10×10−30m3 Hence, mass of one virus = density × volume mass of avogadro number of virus = NA × mass of one virus = molecular weight of the virus.

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