A plant virus is found to consist of uniform cylindrical particle of `150 Å` in diameter 5000 Å long. The specific volume of the virus is 0.75 `mLg^(-1)`. If the virus is considered to be a single particle, find its molar mass.

Explanation

To find the molar mass of the plant virus, we will follow these steps:

### Step 1: Calculate the Volume of the Cylindrical Virus Particle
The formula for the volume \( V \) of a cylinder is given by:
\[ V = \pi r^2 h \]
where \( r \) is the radius and \( h \) is the height (or length) of the cylinder.

**Given:**
- Diameter = 150 Å
- Length = 5000 Å

First, we need to convert these measurements from angstroms to centimeters:
- 1 Å = \( 10^{-8} \) cm
- Diameter = 150 Å = \( 150 \times 10^{-8} \) cm
- Radius \( r = \frac{Diameter}{2} = \frac{150 \times 10^{-8}}{2} = 75 \times 10^{-8} \) cm
- Length \( h = 5000 \) Å = \( 5000 \times 10^{-8} \) cm

Now substituting the values into the volume formula:
\[ V = \pi (75 \times 10^{-8})^2 (5000 \times 10^{-8}) \]
Calculating this gives:
\[ V \approx 8.839 \times 10^{-17} \text{ cm}^3 \]

### Step 2: Calculate the Mass of the Virus Particle
Using the specific volume \( V_s \) to find the mass \( m \):
\[ V_s = \frac{V}{m} \implies m = \frac{V}{V_s} \]
**Given:**
- Specific volume \( V_s = 0.75 \text{ mL/g} = 0.75 \text{ cm}^3/\text{g} \)

Now substituting the values:
\[ m = \frac{8.839 \times 10^{-17} \text{ cm}^3}{0.75 \text{ cm}^3/\text{g}} \approx 11.78 \times 10^{-17} \text{ g} \]

### Step 3: Calculate the Molar Mass of the Virus
The molar mass \( M \) is calculated using the mass of one particle and Avogadro's number \( N_A \):
\[ M = m \times N_A \]
**Given:**
- Avogadro's number \( N_A = 6.022 \times 10^{23} \text{ particles/mol} \)

Substituting the values:
\[ M = (11.78 \times 10^{-17} \text{ g}) \times (6.022 \times 10^{23} \text{ particles/mol}) \approx 7.098 \times 10^{7} \text{ g/mol} \]

### Final Answer
The molar mass of the plant virus is approximately:
\[ \boxed{7.098 \times 10^{7} \text{ g/mol}} \]

---