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. (a)`3.08xx10^(3) kg//mol`
B. (b)`15.4 kg//mol`
C. (c )`1.54xx10^(4) kg//mol`
D. (d)`3.08xx10^(4) kg//mol`

Explanation

To find the molecular weight of the cylindrical virus particle, we can follow these steps:

### Step 1: Convert the dimensions from angstroms to centimeters
Given:
- Radius \( r = 7 \, \text{Å} \)
- Length \( L = 10 \, \text{Å} \)

Since \( 1 \, \text{Å} = 10^{-8} \, \text{cm} \):
- Radius in cm:
\[ r = 7 \, \text{Å} = 7 \times 10^{-8} \, \text{cm} \]
- Length in cm:
\[ L = 10 \, \text{Å} = 10 \times 10^{-8} \, \text{cm} \]

### Step 2: Calculate the volume of the cylindrical virus particle
The volume \( V \) of a cylinder is given by the formula:
\[ V = \pi r^2 L \]
Substituting the values:
\[ V = \pi (7 \times 10^{-8} \, \text{cm})^2 (10 \times 10^{-8} \, \text{cm}) \]
Calculating:
\[ V = 3.14 \times (49 \times 10^{-16} \, \text{cm}^2) \times (10 \times 10^{-8} \, \text{cm}) \]
\[ V = 3.14 \times 490 \times 10^{-24} \, \text{cm}^3 \]
\[ V \approx 1540 \times 10^{-24} \, \text{cm}^3 = 1.54 \times 10^{-22} \, \text{cm}^3 \]

### Step 3: Use the specific volume to find the weight of one virus particle
Given specific volume \( = 6.02 \times 10^{-2} \, \text{cm}^3/\text{g} \):
\[ \text{Weight of 1 virus particle} = \frac{\text{Volume of 1 virus particle}}{\text{Specific volume}} \]
Substituting the values:
\[ \text{Weight} = \frac{1.54 \times 10^{-22} \, \text{cm}^3}{6.02 \times 10^{-2} \, \text{cm}^3/\text{g}} \]
Calculating:
\[ \text{Weight} \approx 2.56 \times 10^{-21} \, \text{g} \]

### Step 4: Calculate the molecular weight of the virus
The molecular weight \( M \) is given by:
\[ M = N_A \times \text{Weight of 1 particle} \]
Where \( N_A = 6.02 \times 10^{23} \):
\[ M = 6.02 \times 10^{23} \times 2.56 \times 10^{-21} \, \text{g} \]
Calculating:
\[ M \approx 154 \, \text{g/mol} \]

### Final Answer
The molecular weight of the virus is approximately \( 154 \, \text{g/mol} \).

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