Question

In: Chemistry

0.916 g He, 11.780 g F2, and 17.960 g Ar are placed in a 16.4-L container...

0.916 g He, 11.780 g F2, and 17.960 g Ar are placed in a 16.4-L container at 23 oC. What are the molar concentrations and partial pressures of the gases?
(All values to the nearest 0.0001)

gas Molar concentration Partial pressure
He M atm
F2 M atm
Ar M atm

What is the total pressure in the container?

P =  atm

Solutions

Expert Solution

Molar mass of helium = 4 g/mol

Moles of helium = 0.916/4 = 0.229 mol

Molar concentration of helium = moles of He/ volume of the container

                                                 = 0.229/16.4

                                                 = 0.014 M

From ideal gas law,

PV= nRT

Where,

P = pressure

V= volume = 16.4 L

n = number of moles = 0.229 mol

R = gas constant = 0.082 L atm mol-1K-1

T = temperature = 23 oC = (273 + 23) = 296 K

or, n/V = P/RT

or, 0.229/16.4 = P/ 0.082 x 296

or, P = 0.34 atm

Therefore the partial pressure of helium = 0.34 atm

--------------------------------------------

Molar mass of fluorine = 38 g/mol

Moles of fluorine= 11.780/38 = 0.310 mol

Molar concentration of fluorine= moles of F2/ volume of the container

                                                 = 0.310/16.4

                                                 = 0.019 M

From ideal gas law,

PV= nRT

Where,

P = pressure

V= volume = 16.4 L

n = number of moles = 0.310 mol

R = gas constant = 0.082 L atm mol-1K-1

T = temperature = 23 oC = (273 + 23) = 296 K

or, n/V = P/RT

or, 0.310/16.4 = P/ 0.082 x 296

or, P = 0.46 atm

Therefore the partial pressure of fluorine = 0.46 atm

-------------------------------------------------

Molar mass of argon = 40 g/mol

Moles of argon= 17.96/40 = 0.449 mol

Molar concentration of argon = mole of Ar/ volume of the container

                                               = 0.449/16.4

                                               = 0.027 M

From ideal gas law,

PV= nRT

Where,

P = pressure

V= volume = 16.4 L

n = number of moles = 0.449 mol

R = gas constant = 0.082 L atm mol-1K-1

T = temperature = 23 oC = (273 + 23) = 296 K

or, n/V = P/RT

or, 0.449/16.4 = P/ 0.082 x 296

or, P = 0.66 atm

Therefore the partial pressure of argon = 0.66 atm

---------------

Total pressure in the container = (0.34 + 0.46 + 0.66) = 1.46 atm

queen_honey_blossom answered 2 years ago
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