Question

Specific volume of cylindrical virus particle is $6.02 \times 10^{-2}$ cc/g, whose radius and length are $7 \mathring{A} \, and \, 10\mathring{A}$ respectively. If $N_A = 6.023 \times 10^{23},$ find molecular weight of virus.

• A. $15.4 kg/mol$
• B. $1.54 \times10^4 kg/mol$
• C. $3.08 \times 10^4 kg/mol $
• D. $3.08 \times10^3 kg/mol$

Answer 1

The Correct Option is A

To find the molecular weight of a virus, we need to use the given information about its specific volume and dimensions. Let's break down the solution step-by-step:

1. First, calculate the volume of a cylindrical virus particle. The formula for the volume (V) of a cylinder is: V = \pi r^2 h, where r is the radius and h is the height.
2. Given:
   • Radius (r) = 7 \mathring{A}
   • Height (h) = 10 \mathring{A}
   • Convert these units to centimeters, since 1 \mathring{A} = 10^{-8} \text{cm}
   • Thus, r = 7 \times 10^{-8} \text{cm} and h = 10 \times 10^{-8} \text{cm}
3. Substitute these values into the volume formula: V = \pi (7 \times 10^{-8})^2 (10 \times 10^{-8})
4. Calculate the volume: \begin{align*} V &= \pi (49 \times 10^{-16}) (10 \times 10^{-8}) \\ &= \pi \times 490 \times 10^{-24} \\ &= 1538.6 \times 10^{-24} \text{ cm}^3 \end{align*}
5. The specific volume is given as 6.02 \times 10^{-2} \text{cc/g}. Specific volume relates the volume to the mass, V_{spec} = \frac{V}{m}, where m is mass.
6. Rearrange the formula to find the mass (m) of one virus particle: m = \frac{V}{V_{spec}}
7. Substituting the known values: \begin{align*} m &= \frac{1538.6 \times 10^{-24} \text{ cm}^3}{6.02 \times 10^{-2} \text{cc/g}} \\ &= 25.56 \times 10^{-22} \text{ g/particle} \end{align*}
8. We are also given Avogadro's number (N_A = 6.023 \times 10^{23} \text{ particles/mol}).
9. To find the molecular weight of the virus: \text{Molecular weight} = m \times N_A
10. Substitute the values: \begin{align*} \text{Molecular weight} &= 25.56 \times 10^{-22} \text{ g/particle} \times 6.023 \times 10^{23} \text{ particles/mol} \\ &= 15.4 \times 10^{3} \text{ g/mol} \\ &= 15.4 \text{ kg/mol} \end{align*}

Therefore, the molecular weight of the virus is 15.4 \text{ kg/mol}.