Question

A 226.4-l cylinder contains 65.5% he(g) and 34.5% kr(g) by mass at 27.0°c and 1.40 atm total pressure. what is the mass of he in this container?

Answer 1

Answer : The mass of helium (He) in this container is, 0.2 grams

Solution :

First we have to calculate the moles of gas present in the container.

Using ideal gas equation,

$PV=nRT\\\\n=\frac{RT}{PV}$

where,

P = pressure of the gas present in the container = 1.40 atm

V = volume of the gas present in the container = 226.4 L

R = gas constant = 0.0821 Latm/moleK

T = temperature of the gas = $27^oC=273+27=300K$

n = moles of gas present in the container = ?

Now put all the given values in the above equation, we get the moles of gas present in the container.

$n=\frac{(0.0821Latm/moleK)\times (300K)}{(226.4L)\times (1.40atm)}=0.077moles$

Now we have to calculate the moles of helium.

$\text{Moles of helium}=0.077\times \frac{65.5}{100}=0.050mole$

Now we have to calculate the mass of helium present in the container.

$\text{Mass of helium}=\text{Moles of helium}\times \text{Molar mass of helium}=0.050moles\times 4g/mole=0.2g$

Therefore, the mass of helium (He) in this container is, 0.2 grams

Answer 2

First, we determine the number of moles of gas present using the ideal gas equation.

PV = nRT
n = PV / RT
n = (1.4 * 226.4) / (0.082 *(27 + 273.15))
n = 12.88

Next, we use the given percentages to find the moles of helium present
Moles of helium = 0.655 * 12.88
Moles of helium = 8.44

Next, we use the formula:
Mass = moles * molar mass

Mass of helium = 8.44 * 4

Mass of helium = 33.76 grams