Question

A mixture of methane and krypton gases is maintained in a 8.99 L flask at a pressure of 1.68 atm and a temperature of 46 °C. If the gas mixture contains 4.88 grams of methane, the number of grams of krypton in the mixture is :

Answer 1

22.81 g

Step-by-step explanation:

Given that:

Pressure = 1.68 atm

Temperature = 46 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (46 + 273.15) K = 319.15 K

Volume = 8.99 L

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

1.68 atm × 8.99 L = n × 0.0821 L.atm/K.mol × 319.15 K

⇒n = 0.5764 moles

Given that :

Amount of methane = 4.88 g

Molar mass = 16.04 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (4.88\ g)/(16.04\ g/mol)`

`Moles= 0.3042\ mol`

Moles of Krypton = Total moles - Moles of methane = 0.5764 - 0.3042 moles = 0.2722 moles

Also, Molar mass of krypton = 83.798 g/mol

So,

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`0.2722\ moles=(Mass)/(83.798\ g/mol)`

Mass of krypton = 22.81 g