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## Q. 5.4

Ammonia gas is used as a refrigerant in food processing and storage industries. Calculate the volume (in liters) occupied by 7.40 g of $NH_{3}$ at STP.

Strategy What is the volume of 1 mole of an ideal gas at STP? How many moles are there in 7.40 g of $NH_{3}$?

## Verified Solution

Recognizing that 1 mole of an ideal gas occupies 22.41 L at STP and using the molar mass of $NH_{3}$ (17.03 g), we write the sequence of conversions as

grams of  $NH_{3}$ → moles of  $NH_{3}$ → liters of  $NH_{3}$ at STP

so the volume of  $NH_{3}$ is given by

$V=7.40 \cancel{g NH_{3}}\times \frac{1 \cancel{mol NH_{3}}}{17.03 \cancel{g NH_{3}}} \times \frac{22.41 L}{1 \cancel{mol NH_{3}}}$

=9.74  L

It is often true in chemistry, particularly in gas law calculations, that a problem can be solved in more than one way. Here the problem can also be solved by first converting 7.40 g of $NH_{3}$ to number of moles of   $NH_{3}$, and then applying the ideal gas equation (V = nRT⁄P). Try it.

Check Because 7.40 g of  $NH_{3}$ is smaller than its molar mass, its volume at STP should be smaller than 22.41 L. Therefore, the answer is reasonable.