Specific volume of cylindrical virus particle is `6.02xx10^(-2) c c//g` whose radius and length `7 Å` and `10 Å` respectively. If `N_(A)=6.02xx10^(23)`, find molecular weight of virus:

Options

A. `15.4 kg//mol`
B. `1.54xx10^(4)kg//mol`
C. `3.08xx10^(4)kg//mol`
D. `3.08xx10^(3)kg//mol`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Convert Angstroms to Centimeters
Given:
- Radius (R) = 7 Å
- Length (L) = 10 Å

1 Å = \(10^{-8}\) cm

So,
- Radius in cm = \(7 \times 10^{-8}\) cm
- Length in cm = \(10 \times 10^{-8}\) cm

### Step 2: Calculate the Volume of the Cylindrical Virus
The formula for the volume (V) of a cylinder is:
\[ V = \pi R^2 L \]

Substituting the values:
- \( R = 7 \times 10^{-8} \) cm
- \( L = 10 \times 10^{-8} \) cm

\[ V = \pi (7 \times 10^{-8})^2 (10 \times 10^{-8}) \]
\[ V = \pi (49 \times 10^{-16}) (10 \times 10^{-8}) \]
\[ V = \pi (49 \times 10^{-24}) \]
\[ V = 154 \times 10^{-23} \text{ cc} \] (using \(\pi \approx \frac{22}{7}\))

### Step 3: Calculate the Weight of One Virus Particle
The specific volume (SV) is given as:
\[ SV = 6.02 \times 10^{-2} \text{ cc/g} \]

Weight (W) of one virus particle can be calculated using:
\[ W = \frac{V}{SV} \]

Substituting the values:
\[ W = \frac{154 \times 10^{-23}}{6.02 \times 10^{-2}} \]

### Step 4: Calculate the Molecular Weight of the Virus
The molecular weight (MW) can be calculated using Avogadro's number (\(N_A = 6.02 \times 10^{23}\)):
\[ MW = W \times N_A \]

Substituting the values:
\[ MW = \left(\frac{154 \times 10^{-23}}{6.02 \times 10^{-2}}\right) \times (6.02 \times 10^{23}) \]

### Step 5: Simplify the Expression
\[ MW = \frac{154 \times 10^{-23} \times 6.02 \times 10^{23}}{6.02 \times 10^{-2}} \]
\[ MW = \frac{154 \times 6.02}{6.02} \times 10^{0} \]
\[ MW = 154 \times 10^{0} \]
\[ MW = 154 \text{ grams/mole} \]

### Step 6: Convert to Kilograms per Mole
Since the answer is required in kilograms per mole:
\[ MW = \frac{154}{1000} = 0.154 \text{ kg/mole} \]

### Final Answer
The molecular weight of the virus is approximately **15.4 kg/mole**.

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