The density of ammonia (NH3, molar mass = 17.0 g/mol) in a 4.32 L container at 837 torr and 45.0 °C is g/L. (Assume that ammonia is an ideal gas; R = 0.08206 atm-L/mol-K; 1 atm = 760 torr)
0.432
3.86
0.0422
0.717
0.194′
The correct answer and explanation is :
To determine the density of ammonia in the given conditions, we will use the ideal gas law formula and its density variation:
PV=nRTPV = nRT
Rearranging for density: d=PMRTd = \frac{PM}{RT} where:
- PP = pressure in atm
- MM = molar mass (g/mol)
- RR = ideal gas constant (0.08206 atm·L/mol·K)
- TT = temperature in Kelvin
Given:
- P=837P = 837 torr = 837760\frac{837}{760} atm
- M=17.0M = 17.0 g/mol
- T=45.0+273.15T = 45.0 + 273.15 K
- R=0.08206R = 0.08206