Question

1. What is the molar volume of 18.40 g of methane in a volume of 0.9466 L? 2. How many grams of ammonia are present in 5.87 L of gas if the molar volume is 26.05 L/mol? If this gas has a temperature of 19.000C what is its pressure in atmospheres? 3. Show mathematically how the molar volume of a gas is related to the density of that gas.

Answer 1

Question 1.

Molar Volume = volume of gas / moles of gas

for methane

Molar Mass CH4 = 16.0425 g/mol

moles of CH4 = mass/M = 18.40/16.0425= 1.146953 mol of CH4

Now,

Vmolar = 0.9466 L /1.146953 mol = 0.8253171 Liters per mol

Question 2.

find moles of ammonia; given Vmolar = 2605 L / mol

Vmolar = volume / moles

moles = Volume / Vmolar = (5.87 L )/(26.05 L /mol)= 0.225335 moles

if T = 19°C = 19+273.15 = 292.15 K; P = x?; R = 0.082 Latm/molK

Apply Ideal gas law:

PV = nRT

where

P = absolute pressure

V = total volume of gas

n = moles of gas

T = absolute Tmperature

R = ideal gas constant

PV = nRT

P = nRT/V

P =(0.225335)(0.082)(292.15)/(5.87)

P = 0.91962 atm

Question 3.

Apply ideal gas law

PV = nRT

mol = mass/MW ---> n = m/MW

PV = (m/MW) * RT

note that

Density = mass / vol --> D = m/V

P = (m/MW) * RT / V

P = (m/V) *RT/MW

P = D*RT/MW

D = P*MW/(RT)

Note that:

Molar volume --> Volume of gas / moles of gas

PV = nRT

V/n = RT/P

and we know,

mass = moles*Molar mas --> m = n*MW --> n = m/MW

V/(m/MW) = RT/P

V*MW/m = RT/P

Since Density = m/V then

MW/D = RT/P

D = P*MW/(RT)

which is exactly the SAME equation we got previously