If the `v_(rms)` is `30R^(1//2)` at `27^(@)C` then calculate the molar mass of gas in kilogram.

Options

A. 0.02 kg/mol
B. 0.001 kg/mol
C. 0.003 kg/mol
D. 1 kg/mol

Explanation

To solve the problem step by step, we need to use the formula for the root mean square velocity (\(v_{rms}\)) of a gas, which is given by:

\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]

where:
- \(v_{rms}\) = root mean square velocity
- \(R\) = universal gas constant
- \(T\) = absolute temperature in Kelvin
- \(M\) = molar mass of the gas in kg/mol

### Step 1: Convert the temperature from Celsius to Kelvin
The temperature given is \(27^\circ C\). To convert this to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273.15 \]

So,

\[ T = 27 + 273.15 = 300.15 \, K \approx 300 \, K \]

### Step 2: Set up the equation for \(v_{rms}\)
We know from the problem that:

\[ v_{rms} = 30\sqrt{R} \]

Substituting this into the \(v_{rms}\) formula gives us:

\[ 30\sqrt{R} = \sqrt{\frac{3RT}{M}} \]

### Step 3: Square both sides to eliminate the square root
Squaring both sides results in:

\[ (30\sqrt{R})^2 = \frac{3RT}{M} \]

This simplifies to:

\[ 900R = \frac{3RT}{M} \]

### Step 4: Rearrange the equation to solve for \(M\)
Now, we can rearrange the equation to isolate \(M\):

\[ M = \frac{3RT}{900R} \]

The \(R\) in the numerator and denominator cancels out:

\[ M = \frac{3T}{900} \]

### Step 5: Substitute the value of \(T\)
Now, substituting \(T = 300 \, K\):

\[ M = \frac{3 \times 300}{900} \]

This simplifies to:

\[ M = \frac{900}{900} = 1 \, \text{g/mol} \]

### Step 6: Convert grams per mole to kilograms per mole
To convert the molar mass from grams per mole to kilograms per mole, we divide by 1000:

\[ M = 1 \, \text{g/mol} = \frac{1}{1000} \, \text{kg/mol} = 0.001 \, \text{kg/mol} \]

### Final Answer
Thus, the molar mass of the gas is:

\[ \boxed{0.001 \, \text{kg/mol}} \]