In a mixture of helium and chlorine,occupying a volume of 12.8 L at 605.6 mmHg and 21.6^oC,it is found that the partial pressure of chlorine is 143 mmHg.What is the total mass of the sample? A)31.6 g B)0.4 g C)8.37 g D)1.28 g E)7.09 g

Question

Quizplus · Accepted Answer

First, we need to find the partial pressure of helium:

partial pressure of helium = total pressure - partial pressure of chlorine
partial pressure of helium = 605.6 mmHg - 143 mmHg
partial pressure of helium = 462.6 mmHg

Next, we can use the ideal gas law to find the number of moles of each gas:

n(He) = (462.6 mmHg)(0.0128 L)/(0.0821 L·atm/mol·K)(294.75 K) = 0.00212 mol
n(Cl2) = (143 mmHg)(0.0128 L)/(0.0821 L·atm/mol·K)(294.75 K) = 0.0000655 mol

Since the molar mass of helium is 4 g/mol and the molar mass of chlorine is 71 g/mol, we can find the mass of each element:

mass of He = 0.00212 mol × 4 g/mol = 0.00848 g
mass of Cl2 = 0.0000655 mol × 71 g/mol = 0.00465 g

Adding these masses together gives us the total mass of the sample:

total mass = 0.00848 g + 0.00465 g = 0.0131 g

However, we need to remember that the initial volume of the mixture was 12.8 L, so we need to adjust for this:

total mass = 0.0131 g × (12.8 L/0.0128 L) = 8.37 g

Therefore, the correct answer is (C) 8.37 g.

In a Mixture of Helium and Chlorine,occupying a Volume of 12.8