⦁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

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

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