Chapter 6 - Gases - Example 6-4 - Calculating a Gas Volume with the Ideal Gas Equation - Page 205: Practice Example A

24.4 L

$n= 20.2\,g\,NH_{3}\times\frac{1\,mol\,NH_{3}}{17.031\,g\,NH_{3}}=1.186\,mol$ $T= (-25+273)\,K=248\,K$ $R=0.08206\,atm\,L\,mol^{-1}K^{-1}$ $P=752\,mmHg\times\frac{1\,atm}{760\, mmHg}$ $=0.98947\,atm$ Using the ideal gas law, we have $V=\frac{nRT}{P}=\frac{1.186\,mol\times0.08206\,atm\,L\,mol^{-1}K^{-1}\times248\,K}{0.98947\,atm}$ $=24.4\,L$