Final answer:
To calculate the vapor density of phosphine at 300K and 2.5atm, we can use the ideal gas law. The ideal gas law is given by PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Given that phosphine shows 40% dissociation in equilibrium, we can assume that 60% of the initial amount of phosphine remains undissociated. Therefore, the vapor density is 2788 g/m³.
option d is the correct
Step-by-step explanation:
To calculate the vapor density of phosphine at 300K and 2.5atm, we can use the ideal gas law. The ideal gas law is given by PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. We can rearrange the equation to solve for n/V: (n/V) = P/RT.
Given that phosphine shows 40% dissociation in equilibrium, we can assume that 60% of the initial amount of phosphine remains undissociated. Therefore, the pressure due to undissociated phosphine is 0.6 * 2.5atm = 1.5atm.
Substituting the values into the equation, we get (n/V) = (1.5atm) / (0.0821 L*atm/(mol*K) * 300K) = 0.082 mol/L.
The molar mass of phosphine is 34g/mol, so the molecular weight of phosphine is 0.082 mol/L * 34 g/mol = 2.788 g/L. The vapor density is the mass of the gas divided by its volume, so the vapor density is 2.788 g/L.
Since the question asks for the vapor density in g/m³, we need to convert the units. There are 1000 L in a m³, so the vapor density in g/m³ is 2.788 g/L * (1000 L/m³) = 2788 g/m³.