Specific volume of cylindrical virus particle is $6 .02 \times 10^{- 2} cc / g$ whose radius and length $7 A^{0} and 10 A^{0}$ respectively. If $N_{A} = 6.02 \times 10^{23},$ find molecular weight of virus

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$1 .54 \times 10^{4} kg / mol$

$3 .08 \times 10^{3} kg / mol$

15.4 kg/mol

$3 .08 \times 10^{4} kg / mol$

answer is B.

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Detailed Solution

Specific volume (volume of g) of cylindrical virus particle

$= 6 .02 \times 10^{- 2} cc / g$

Radius of virus (r) = $7 A^{0} = 7 \times 10^{- 8} c m$

$\begin{array}{l} Length of virus = 10 \times 10^{- 8} cm \\ Volume of virus \\ {πr}^{2} L = \frac{22}{7} \times {(7 \times 10^{- 8})}^{2} \times 10 \times 10^{- 8} = 154 \times 10^{- 23} cc \\ Wt . of one virus particle = \frac{volume}{specific volume} \\ ∴ Mol . wt . of virus = wt . of N_{A} particle \\ = \frac{154 \times 10^{- 23}}{6 .02 \times 10^{- 2}} \times 6 .02 \times 10^{23} \\ = 15400 g / mol = 15 .4 kg / mol \end{array}$