Question

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

Answer 1

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`