Question

1.) The total pressure (P_T) os a sample of a gas collected over water at 13.0^oC is 0.588 atm. Determine the partial pressure (P_i) of the gas.

2.) Rearrange the ideal gas equation to solve for density (g/v). Hint: you will need to make a substitution for moles "n"

3.) Use your quation from (z) and the result from (l) to calculate the molecular weight of a gas if 303 ml of the gas, collected at 28^oC, weighed 0.347g.

Answer 1

Answer – 1) Given, total pressure = 0.588 atm at 13.0^oC

We know at 13.0^oC the water vapour pressure is 11.2 torr

Now we need to convert the partial pressure of water in torr to atm

We know

760 torr = 1 atm

So, 11.2 torr = ?

= 0.0147 atm

We know the Daltons partial pressure law

Total pressure = sum of the partial pressure

       0.588 atm = partial pressure of the gas + 0.0147 atm

So, partial pressure of the gas = 0.588 – 0.0147 atm

                                                = 0.573 atm

2) We know the Ideal gas law is

PV =nRT

We know, n = moles and we can calculate moles from the formula

Moles = mass / molar mass

            = g/g.mol^-1

So by substituting the value of n in the Idea gas law

PV = mass/Molar mass * RT

PV/RT = mass / molar mass

PV*molar mass / RT = mass

So, P*molar mass / RT = mass / V

We know, density = mass / volume

So, density = PM/RT

Where, M = molar mass.

3) Given, total pressure of gas = 0.588 atm , volume = 303 mL = 0.303 L

T = 28^oC + 273 = 301 K , mass = 0.347 g

At 28^oC the water vapor pressue = 28.4 torr

                                                     = 28.4 torr / 760 = 0.0374 atm

Partial pressure of the gas = 0.588 – 0.0374 atm

                                            = 0.551 atm

   Now using the formula

Density = PM/RT

We know, density = mass / volume

0.347 g / 0.303 L = 0.551 atm * M /0.0821 L.atn.mol^-1.K^-1 * 301 K

1.14 = 0.551 atm * M / 24.71

So, M = 1.14*24.71 / 0.551

           = 51.396 g/mol