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
Kai
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#

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