Question

what volume would 120.0g of nitrogen gas occupy at 59.50 c and 0.566 atm

Answer 1

59.50^oC = 59.50 + 273.15 = 332.65 K

$n(N_2)=\frac{120.0\ g}{28.01\ g/mol}=4.284\ mol$

$R=0.08206\frac{L\cdot{atm}}{mol\cdot{K}}$

According to the ideal gas law,

$PV=nRT$ ;

$V=\frac{nRT}{P}=\frac{4.284\ mol\times0.08206\frac{L\cdot{atm}}{mol\cdot{K}}\times332.65\ K}{0.566\ atm}=207\ L$

Answer: 207 L