Question

the density of a gaseous fluoride of phosphorus is 3.93 g/l at stp. calculate the molar mass of the fluoride and determine its molecular formula

Answer 1

Data given : density of fluoride of phosphorous = 3.93 g/lite at stp

At stp condition, Temperature = 0 oC = 273.15 K , P = 1 atm

To solve this we make use of ideal gas law

Ideal gas law is given by,

P V = N R T

Where,

P = pressure

V= volume

N= Number of moles

R = Ideal gas constant = 0.0821 Litre atm mole^-1 k^-1

T= temperature

Now we will write ideal gas equation in terms of density

P V = (W/Mw) R T

Where, W = weight of fluoride of phosphor, Mw =molar mass of fluoride of phosphor

P Mw =(W/V) R T

but density (Rhou) = W/V

P Mw = Rhou * R*T

Now we will substitute all the given data in above equation to obtain the value of molar mass

1 * Mw = 3.93 * 0.0821 * 273.15

Mw = 88.1326 gram/mole

Molar mass of fluoride of phosphor is 88.1326 gram/mole

Let molecular formula of fluoride be PF_x

Hence molecular formula, in terms of x, will be = 31+ 19x

Since we already have obtained the value of molar mass we will equate it to "31+19x" to obtain the value of x

88.1326 = 31+ 19x

from above equation , x=3

Hence molecular formula of Fluoride of phosphor is PF₃