Question

What is the density of NH3 at 800 torr and 25 degree celsius?

Answer 1

To solve this question, we need to use the following formula:

`\rho\text{ = }((MM_(NH_3))(P))/((R)(T))`

where:

d = density

MMNH3 = molar mass of NH3 = 17 g/mol

P = pressure = 800 Torr = 1.05263 atm (just divide by 760)

T = 25 °C = 273.15 + 25 = 298.15 K

R = 0.082 atm

So:

d = 17 * 1.05263/0.082 * 298.15

d = 17.8947/24.4483

d = 0.7319 g/L

Answer: density = 0.7319 g/L