A 10.0-g sample of krypton has a temperature of \( 25^{\circ} \mathrm{C} \) at 577 mmHg . What is the volume, in milliiiters, of the krypton gas? Express your answer to three significant figures and include the appropriate units.

To find the volume of the krypton gas, we can use the Ideal Gas Law, which is given by the equation \( PV = nRT \). First, we need to determine the number of moles of krypton in the sample. The molar mass of krypton (Kr) is approximately \( 83.80 \, \text{g/mol} \). \[ n = \frac{mass}{molar \, mass} = \frac{10.0 \, g}{83.80 \, g/mol} \approx 0.119 mol \] Next, we convert the temperature from Celsius to Kelvin: \[ T(K) = 25^{\circ}C + 273.15 = 298.15 \, K \] We also need to convert pressure from mmHg to atm: \[ P(atm) = \frac{577 \, mmHg}{760 \, mmHg/atm} \approx 0.759 \, atm \] Now we can rearrange the Ideal Gas Law to solve for volume \( V \): \[ V = \frac{nRT}{P} \] Substituting in the values: - \( n \approx 0.119 \, mol \) - \( R = 0.0821 \, L \cdot atm/(mol \cdot K) \) - \( T \approx 298.15 \, K \) - \( P \approx 0.759 \, atm \) \[ V = \frac{(0.119 \, mol)(0.0821 \, L \cdot atm/(mol \cdot K))(298.15 \, K)}{0.759 \, atm} \approx 3.66 \, L \] To convert liters to milliliters: \[ V \approx 3.66 \, L \times 1000 \, mL/L = 3660 \, mL \] Thus, the volume of the krypton gas is approximately \( 3660 \, \text{mL} \). So, your final answer, expressed in three significant figures and with appropriate units, is: **3660 mL**