A10 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 this temperature, how many grams of liquid remain after the establishment of the liquid-vapor equilibrium?

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!

i think that the answer is c

the correct answer would be b) volume versus temperature, because they are also directly proportional.