How many grams are contained in a 0.183 mol sample of ammonium phosphate?
1 answer:
You might be interested in
Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.
Explanation : Given,
Vapor presume of 1‑Propanol = 20.9 torr
Vapor presume of 2‑Propanol = 45.2 torr
Mole fraction of 1‑Propanol = 0.540
Mole fraction of 2‑Propanol = 1-0.540 = 0.46
First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.
where,
= partial vapor pressure of 1‑Propanol
= vapor pressure of pure substance 1‑Propanol
= mole fraction of 1‑Propanol
and,
where,
= partial vapor pressure of 2‑Propanol
= vapor pressure of pure substance 2‑Propanol
= mole fraction of 2‑Propanol
Thus, total pressure = 11.3 + 20.8 = 32.1 torr
Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.
and,
Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.