You have 894 mL of chlorine trifluoride gas at 772 mmHg and 28°C. What is the mass (in g) of the sample?

1 Answer
Mar 7, 2018

The mass of ClF_3 in the sample is 3.41 g.

Explanation:

To solve this problem, we need to use the ideal gas law:

PV=nRT

P is pressure in atmospheres, V is volume in liters, n is moles, R is the universal gas constant, and T is the temperature in Kelvins.

First, we will need to convert each of the units.

894color(white)(l)-:1000=0.894 color(white)(l)L

(772color(white)(l)mmHg)/1times(1color(white)(l)atm)/(760color(white)(l)mmHg)=1.02 atm

28+273=301color(white)(l)K

Now, plug the above values into the equation and solve for moles.

(1.02)(0.894)=(n)(0.08206)(301)

n=((1.02)(0.894))/((0.08206)(301))=0.0369color(white)(l)molcolor(white)(l)ClF_3

Finally convert moles of ClF_3 to grams:

(0.0369color(white)(l)molcolor(white)(l)ClF_3)/1times(92.448color(white)(l)gcolor(white)(l)ClF_3)/(1color(white)(l)molcolor(white)(l)ClF_3)=3.41color(white)(l)gcolor(white)(l)ClF_3