Question

⦁answer Calculate the density in g/L of 478 mL of krypton at 47° C and 671 mm Hg. ⦁ Determine the molar mass of a gas that has a density of 2.18 g/L at 66°C and 720 mm Hg. ⦁ Find the mass of 5.60 L O2 at 1.75 atm and 250.0 K

Answer 1

Answer:

Explanation:

First of all we have to convert our variables to work with equal terms in our ideal gas equation.

PV = nRT

P = pressure = 671mmHg = 0.882atm

Note: (1mmHg = 0.001315atm)

V = volume of the gas = 478cm³

n = no. Of moles =?

T = 47°C = 320K

R = ideal gas constant = 0.08205

cm³atm/mol.k

PV = nRT

n = PV / RT

n = (0.882*478) / (0.08205*320)

n = 16.057 moles

n = mass / molarmass

Molar mass of krypton = 85.798 u

Mass = no. Moles * molar mass

Mass = (16.057 * 85.798) = 1377.66g

Density (ρ) = mass / volume

ρ = m / v = (1377.66 / 478) = 2.88g/cm³

(B) = ρ = 2.18g/L

T = 66°C = 339K

P = 720mmHg = 0.9468atm

R = 0.082 atmcm³/Kmol

density = mass / volume

ρ = m / v ....

v = m/ρ

PV = nRT

n = mass / molarmass

P. (m/ρ) = (m/M) * RT

P / ρ = RT /M

M = (ρ *R*T ) / P

M = (2.18 * 0.082 * 339) / 0.9468

M = 64.00g/mol

(C)

PV = nRT

Mass = ?

V = 5.60 cm³

P = 1.75 atm

T = 250K

Molar mass = 16.0

R = 0.082 cm³atm/Kmol

PV = nRT

n = PV / RT

n = (1.75*5.60) / (0.082*250)

n = 9.8 / 20.5

n = 0.4780 moles

n = mass / molar mass

Mass = no. Of moles * molar mass

Mass = 0.4780 * 16.0

Mass = 7.647g

Answer 2

Answer:

The correct answers are:

- Krypton: density= 2.8 g/L

- Molar Mass= 63.99 g/mol

- Mass of O₂= 15.29 g

Explanation:

The general equation of an ideal gas is the folllowing:

P x V = n x R x T

Where: P= pressure (in atm), V= volume; n= number of moles, R= gas constant (0,082 L.atm/K.mol) and T= temperature (in K).

For krypton:

P= 671 mmHg = 0,882 atm

V= 478 ml x 1000 ml/1 L= 0,478 L

T= 47ºC= 320 K

MM= 83.8 g/mol (from Periodic Table, Kr is an inert gas so it is a monoatomic gas)

P x V = n x R x T

Since the number of moles of a compound can be calculated by dividing the mass of compound (m) into its molar mass (MM):

n= m/MM

We can replace the expression in the first equation to obtain:

$P x V= \frac{m}{MM} x R x T$

m/V= $\frac{P x MM}{R x T}$

Density (d) is equal to the mass per volume (m/V), so we can directly calculate the density:

d= m/V= \frac{P x MM}{R x T}=

           = (0.882 atm x 83.8 g/mol)/(0.082 L.atm/K.mol x 320 K)

           = 2.81 g/L

For the gas:

d= 2.18 g/L

T= 66ºC= 339 K

P= 720 mmHg= 0.947 atm

d= \frac{P x MM}{R x T}

⇒MM = $\frac{dx R x T}{P}$

         = (2.18 g/L x 0.082 L.atm/K.mol x 339 K)/(0.947 atm)

         = 63.99 g/mol ≅ 64 g/mol

For the O₂:

V= 5.60 L

P= 1.75 atm

T= 250 K

MM(O₂) = 2 x Atomic Mass O= 2 x 16 g/mol= 32 g/mol

We can use the second equation:

P x V= \frac{m}{MM} x R x T

⇒  m = $\frac{P x V x MM}{R x T}$= (1.75 atm x 5.6 L x 32 g/mol)/(0.082 L.atm/K.mol x 250 K)

                         = 15.29 g≅ 16 g