According to Newton's formula, the speed of sound in air at STP is
(Take the mass of 1 mole of air is `29 xx 10^(-3)` kg)

Options

A. `250ms^(-1)`
B. `260ms^(-1)`
C. `270ms^(-1)`
D. `280ms^(-1)`

Explanation

To find the speed of sound in air at Standard Temperature and Pressure (STP) using Newton's formula, we will follow these steps:

### Step 1: Understand the given data
We are given:
- Mass of one mole of air = \( 29 \times 10^{-3} \) kg
- Volume of one mole of air at STP = \( 24.2 \) liters = \( 24.2 \times 10^{-3} \) m³ (since \( 1 \) liter = \( 10^{-3} \) m³)

### Step 2: Calculate the density of air
The density (\( \rho \)) of air can be calculated using the formula:
\[ \rho = \frac{\text{mass}}{\text{volume}} \]
Substituting the values:
\[ \rho = \frac{29 \times 10^{-3} \text{ kg}}{24.2 \times 10^{-3} \text{ m}^3} \]
Calculating this gives:
\[ \rho = \frac{29}{24.2} \text{ kg/m}^3 \approx 1.20 \text{ kg/m}^3 \]

### Step 3: Use the pressure at STP
At STP, the pressure (\( P \)) is given as:
\[ P = 1.01 \times 10^{5} \text{ N/m}^2 \]

### Step 4: Apply Newton's formula for the speed of sound
According to Newton's formula, the speed of sound (\( v \)) in a medium is given by:
\[ v = \sqrt{\frac{P}{\rho}} \]
Substituting the values we have:
\[ v = \sqrt{\frac{1.01 \times 10^{5} \text{ N/m}^2}{1.20 \text{ kg/m}^3}} \]

### Step 5: Calculate the speed of sound
Calculating the above expression:
\[ v = \sqrt{\frac{1.01 \times 10^{5}}{1.20}} \approx \sqrt{84166.67} \approx 290.15 \text{ m/s} \]

### Final Answer
Thus, the speed of sound in air at STP is approximately \( 290 \text{ m/s} \).

---