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Step-by-Step Solution
Step 1: Identify the Given Data
• Specific volume of the virus (volume occupied by 1 g of virus),
$6.02 \times 10^{-2}\,\text{cm}^3/\text{g}$.
• Radius of the cylindrical virus, $r = 7 \times 10^{-8}\,\text{cm}$.
• Length of the cylindrical virus, $l = 10 \times 10^{-8}\,\text{cm} = 1 \times 10^{-7}\,\text{cm}$.
• Avogadro's number, $N_A = 6.02 \times 10^{23}\,\text{particles/mol}$.
Step 2: Formula for Volume of a Cylindrical Virus Particle
The virus is modeled as a cylinder of radius $r$ and length $l$. Hence,
$$
\text{Volume of one virus particle} = \pi r^2 \times l
$$
Step 3: Calculate the Volume of One Virus Particle
Substitute $r = 7 \times 10^{-8}\,\text{cm}$ and $l = 1 \times 10^{-7}\,\text{cm}$ into the formula. Using $\pi \approx \frac{22}{7}$ for simplicity:
$$
\text{Volume}
= \frac{22}{7} \times (7 \times 10^{-8}\,\text{cm})^2 \times (1 \times 10^{-7}\,\text{cm})
$$
First, compute $r^2$:
$$
r^2 = (7 \times 10^{-8}\,\text{cm})^2 = 49 \times 10^{-16}\,\text{cm}^2 = 4.9 \times 10^{-15}\,\text{cm}^2
$$
Then multiply by $l$:
$$
4.9 \times 10^{-15}\,\text{cm}^2 \times 1 \times 10^{-7}\,\text{cm}
= 4.9 \times 10^{-22}\,\text{cm}^3
$$
Finally multiply by $\pi \approx 3.14$ (or $22/7$):
$$
\text{Volume}
\approx 3.14 \times 4.9 \times 10^{-22}\,\text{cm}^3
\,\approx\, 1.54 \times 10^{-21}\,\text{cm}^3
$$
(In the given solution, it is expressed as $154 \times 10^{-23}\,\text{cm}^3$, which is the same as $1.54 \times 10^{-21}\,\text{cm}^3$.)
Step 4: Find the Mass of One Virus Particle
The virus has a specific volume of $6.02 \times 10^{-2}\,\text{cm}^3/\text{g}$.
This means every 1 gram of the virus occupies
$6.02 \times 10^{-2}\,\text{cm}^3$.
Therefore, the mass corresponding to a volume $V$ is given by
$$
\text{Mass} = \frac{V}{\text{Specific Volume}}.
$$
Substituting $V = 1.54 \times 10^{-21}\,\text{cm}^3$:
$$
\text{Mass of one virus particle}
= \frac{1.54 \times 10^{-21}\,\text{cm}^3}{6.02 \times 10^{-2}\,\text{cm}^3/\text{g}}
= \frac{1.54}{6.02} \times 10^{-21 + 2} \,\text{g}
= \left(\frac{1.54}{6.02}\right) \times 10^{-19}\,\text{g}.
$$
Step 5: Calculate the Molecular Weight of the Virus
The molecular weight (in grams per mole) is the mass of $N_A$ virus particles.
Since $N_A = 6.02 \times 10^{23}\,\text{particles/mol}$,
$$
\text{Molecular Weight}
= \left[ \text{Mass of one virus particle} \right]
\times N_A.
$$
Hence:
$$
\text{Molecular Weight}
= \left(\frac{1.54 \times 10^{-21}}{6.02 \times 10^{-2}}\right) \times
\left(6.02 \times 10^{23}\right)\,\text{g/mol}.
$$
Numerically, this simplifies to approximately:
$$
\text{Molecular Weight}
\approx 1.54 \times 10^{4}\,\text{g/mol}
= 15.4 \times 10^{3}\,\text{g/mol}
= 15.4\,\text{kg/mol}.
$$
Final Answer
The molecular weight of the virus is $15.4\,\text{kg/mol}$.
