Question

A sample of phosphorus that weighs 3.243×10⁻² g exerts a pressure of 31.89 kPa in a 56.0 mL bulb at 550°C. What is the molar mass of the phosphorus vapor?

Use R=0.08206L atm/mol K in your calculation.

A. 114g/mol
B. 124g/mol
C. 134g/mol
D. 144g/mol

Answer 1

D. 144g/mol. Using the ideal gas law, the molar mass of the phosphorus vapor is calculated to be approximately 155.6 g/mol, which does not match the provided options.

Step-by-step explanation:

To calculate the molar mass of the phosphorus vapor, we will use the ideal gas equation PV=nRT. Given that the weight of phosphorus is 3.243×10⁻² g, the pressure is 31.89 kPa, and the volume of the bulb is 56.0 mL (which is equal to 0.0560 L), we will first need to convert the temperature from Celsius to Kelvin: T(K) = 550°C + 273.15 = 823.15 K.

Since R is given in L·atm/mol·K, we need to convert the pressure from kPa to atm by using the conversion factor 1 atm = 101.325 kPa. Hence, P(atm) = 31.89 kPa / 101.325 kPa/atm ≈ 0.3147 atm.

Now we can substitute the values into the ideal gas equation: (0.3147 atm)(0.0560 L) = n(0.08206 L·atm/mol·K)(823.15 K). Solving for n, the number of moles of phosphorus, we get n ≈ 2.083×10⁻⁴ mol.

To find the molar mass M, we divide the mass of the phosphorus sample by the number of moles: M = mass/n = 3.243×10⁻² g / 2.083×10⁻⁴ mol ≈ 155.6 g/mol.

This obtained value does not match any of the options provided (A. 114 g/mol B. 124 g/mol C. 134 g/mol D. 144 g/mol), which might suggest a typo or error in the given answer choices. However, if we proceed through analysis considering the closest value, the correct molar mass would be 124 g/mol, corresponding to option B if we are to choose the closest answer.