How many liters of HNO₃ gas, measured at 28.0 ºC and 780 torr, are required to prepare 2.30 L of 4.15 M solution of nitric acid?

• A. 56 L
• B. 62 L
• C. 189 L
• D. 230 L
• E. 262 L

Answer: (D)

Verified step by step guidance

1. First, calculate the number of moles of HNO3 needed using the molarity formula: \( M = \frac{n}{V} \), where \( M \) is the molarity, \( n \) is the number of moles, and \( V \) is the volume in liters. Rearrange to find \( n = M \times V \). Substitute \( M = 4.15 \text{ M} \) and \( V = 2.30 \text{ L} \) to find \( n \).
2. Next, use the ideal gas law to find the volume of HNO3 gas required. The ideal gas law is \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.
3. Convert the given temperature from Celsius to Kelvin by using the formula \( T(K) = T(°C) + 273.15 \). For this problem, \( T = 28.0 + 273.15 \).
4. Convert the pressure from torr to atm, since the ideal gas constant \( R \) is typically given in terms of atm. Use the conversion factor \( 1 \text{ atm} = 760 \text{ torr} \). Therefore, \( P = \frac{780}{760} \text{ atm} \).
5. Rearrange the ideal gas law to solve for \( V \): \( V = \frac{nRT}{P} \). Substitute the values for \( n \), \( R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \), \( T \), and \( P \) to find the volume of HNO3 gas required.