Question

Specific volume of cylindrical virus particle is 6.02$\times {10}^{-2}$ cc/gm whose radius and length are 7$\mathring{A}$ and 10$\mathring{A}$ respectively. If $N_A=6.02\times {10}^{23}$, find the molecular weight of virus.

• A. $1.54kg/mol$
• B. $1.54\times {10}^{4}kg/mol$
• C. $3.08\times {10}^{4}kg/mol$
• D. $3.08\times {10}^3$kg/mol

Answer 1

Answer: (A)

$1.54kg/mol$

$\frac{1}{\ell }=6.02\times {10}^{-2}cc/gmr=7\mathring{A}L=10\mathring{A}V=\pi r^{2}L=\pi \times {(7\times {10}^{-8})}^2(10\times {10}^{-8})M=\frac{V}{1/\ell }=\frac{\pi \times 49\times {10}^{-23}}{6.02\times {10}^2}molecularweight=\frac{49\pi \times {10}^{-23}\times 6.02\times {10}^{23}}{6.02\times {10}^2}\left(\frac{49\pi }{100}\right)=1.54kg/mol$