A 10 gram sample of H20 is sealed in a 1350 ml flask at 27°C. Given the fact that water has a vapor pressure of 26.7 mmHg at thi

Question

Aleonysh [2.5K]

3 years ago

10

A 10 gram sample of H20 is sealed in a 1350 ml flask at 27°C. Given the fact that water has a vapor pressure of 26.7 mmHg at thi

s temperature, how many grams of liquid remain after the establishment of the liquid-vapor equilibrium?

Chemistry

1 answer:

Aleksandr-060686 [28] · Answer 1 · 2022-06-16T09:09:43+03:00

Answer:

9.9652g of water

Explanation:

The establishment of the liquid-vapor equilibrium occurs when the vapour of water is equal to vapour pressurem 26.7 mmHg. Using gas law it is possible to know how many moles exert that pressure, thus:

n = PV / RT

Where P is pressure 26,7 mmHg (0.0351atm), V is volume (1.350L), R is gas constant (0.082 atmL/molK) and T is temperature (27°C + 273,15 = 300.15K)

Replacing:

n = 0.0351atm×1.350L / 0.082atmL/molK×300.15K

n = 1.93x10⁻³ moles of water are in gaseous phase. In grams:

1.93x10⁻³ moles × (18.01g / 1mol) = 0.0348g of water

As the initial mass of water was 10g, the mass of water that remains in liquid phase is:

10g - 0.0348g = 9.9652g of water

I hope it helps!