Question

Specific volume of cylindrical virus particles 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 molecular weight of virus.

• A. $15.4$ kg/mol
• B. $1.54\times {10}^4$ kg/mol
• C. $3.08\times {10}^4$ kg/mol
• D. $1.54\times {10}^3$ kg/mol

Answer 1

The correct option is A $15.4$ kg/mol

Solution:- (A) $15.4Kg/mol$

Volume of cylinder $=\pi r^{2}h$

As the virus is cylindrical.

$\therefore$ Volume of one virus $=$ volume of cylinder

As the radius and length of virus are $7\mathring{A}$ and $10\mathring{A}$ respectively.

$\therefore$ Volume of $1$ virus $=\frac{22}{7}\times 7^2\times 10=1540\times {10}^{-24}{cm}^3=1.54\times {10}^{-21}{cm}^3$ $[\because \mathring{1}={10}^{-8}c{m}^3]$

Volume of $1$ mole of virus $=N_A\times$volume of $1$ virus

$\Rightarrow$ Volume of $1$ mole of virus $=6.02\times {10}^{23}\times 1.54\times {10}^{-21}=6.02\times 1.54\times {10}^2{cm}^3$

Given that specific volume of virus is $6.02\times {10}^{-2}{cm}^3/g$

$\therefore \text{Molecular weight}=\frac{\text{volume of 1 mole}}{\text{Specific volume}}=\frac{6.02\times 1.54\times {10}^2}{6.02\times {10}^{-2}}=1.54\times {10}^{4}g/mol=15.4Kg/mol$

Hence, the correct option is A.