Given temperature = [tex] 0^{0}C = (0^{0} + 273 )K = 273 K [/tex]
Pressure = 1.58 ×[tex] 10^{5} Pa [/tex] ×[tex] \frac{1 atm}{1.01325*10^{5}Pa} [/tex]
= 1.56 atm
[tex] Density =\frac{PM}{RT} [/tex]
ρ = [tex] \frac{(1.56 atm)(28.01 g/mol)}{(0.08206 L.atm/mol.K)(273 K)} [/tex]
= 1.95 g/L