Find rms speed of Boron molecules at temperature `27^(@)C`.

Options

A. 824m/s
B. 945m/s
C. 685m/s
D. 258m/s

Explanation

To find the root mean square (rms) speed of boron molecules at a temperature of 27°C, we can follow these steps:

### Step 1: Convert the temperature from Celsius to Kelvin
The formula to convert Celsius to Kelvin is:
\[ T(K) = T(°C) + 273 \]
Given that the temperature is 27°C:
\[ T = 27 + 273 = 300 \, K \]

### Step 2: Identify the molar mass of boron
The molar mass of boron is approximately 11 g/mol. To use it in the rms speed formula, we need to convert it to kilograms per mole:
\[ M = 11 \, g/mol = 11 \times 10^{-3} \, kg/mol \]

### Step 3: Use the rms speed formula
The formula for the rms speed \( v_{rms} \) is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
Where:
- \( R \) is the gas constant, \( R = 8.314 \, J/(K \cdot mol) \)
- \( T \) is the temperature in Kelvin
- \( M \) is the molar mass in kg/mol

### Step 4: Substitute the values into the formula
Now, substituting the values we have:
\[ v_{rms} = \sqrt{\frac{3 \times 8.314 \, J/(K \cdot mol) \times 300 \, K}{11 \times 10^{-3} \, kg/mol}} \]

### Step 5: Calculate the value inside the square root
First, calculate the numerator:
\[ 3 \times 8.314 \times 300 = 7482.6 \, J/mol \]
Now, calculate the denominator:
\[ 11 \times 10^{-3} = 0.011 \, kg/mol \]
Now, divide the numerator by the denominator:
\[ \frac{7482.6}{0.011} = 680,236.36 \, m^2/s^2 \]

### Step 6: Take the square root
Now, take the square root to find the rms speed:
\[ v_{rms} = \sqrt{680236.36} \approx 824.0 \, m/s \]

### Final Answer
The rms speed of boron molecules at 27°C is approximately:
\[ v_{rms} \approx 824 \, m/s \]

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