Question

The density of a certain gaseous fluoride of phosphorus is 3.93 g/L at STP. Calculate the molar mass of this fluoride and determine its molecular formula.

a) Molar mass: 42.04 g/mol, Molecular formula: PF₃
b) Molar mass: 59.98 g/mol, Molecular formula: PF₄
c) Molar mass: 78.00 g/mol, Molecular formula: PF₅
d) Molar mass: 92.94 g/mol, Molecular formula: PF₆

Answer 1

The molar mass of the gaseous fluoride of phosphorus is approximately 46.51 g/mol, and its molecular formula is PF3. The correct answer is A.

Step-by-step explanation:

To calculate the molar mass and molecular formula of the gaseous fluoride of phosphorus, we need to use the ideal gas law equation:

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 in Kelvin.

Given that the pressure is 3.93 g/L, we can convert it to atmospheres (1 atm = 760 mmHg = 101.325 kPa). Therefore, the pressure is approximately 3.93/101.325 = 0.03872 atm.

Using the ideal gas law equation, we can rearrange it to solve for the number of moles:

n = PV/RT

Substituting the values:

n = (0.03872 atm)(0.0560 L)/(0.0821 L⋅atm/mol⋅K)(550 + 273.15 K)

n = 0.00069697 mol (approximately)

Once we have the number of moles, we can calculate the molar mass using the given mass (3.243 × 10^{-2} g) and the number of moles:

Molar mass = mass/number of moles = 3.243 × 10^{-2} g/0.00069697 mol ≈ 46.51 g/mol

So the molar mass of the gaseous fluoride of phosphorus is approximately 46.51 g/mol. The molecular formula would depend on the number and types of atoms present. Based on the molar mass, the molecular formula can be PF3.