Question

What is the K.E. of translational motion of molecules in 15 gram of ammonia gas at `37^(@)C`? (Molecular weight of ammonia is 17.03 and R=8.31 J/mol K)
A. 3404 J
B. 340.3 J
C. 401.5 J
D. 4090 J

Answer 1

Correct Answer - A
`K.E.=(3)/(2)(RT)/(M)m=(1.5xx8.21xx310xx15xx10^(-3))/(17.03xx10^(-3))=3403J`