A 4.7-L sealed bottle containing 0.33 g of liquid ethanol,C₂H₆O, is placed in a refrigerator and reaches equilibrium with its vapor at - 11°C. (c) How much liquid ethanol would be present at 0.0°C? The vapor pressure of ethanol is 10. torr at - 2.3°C and 40. torr at 19°C.

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khuranashilpa76 khuranashilpa76 · Answer 1 · 2022-08-27T19:59:07+02:00

The mass of ethanol present in the liquid phase is 0.15 g. when liquid and vapor ethanol are at equilibrium.

The volume of the bottle = 4.7 L

Mass of ethanol = 0.33 g

Temperature (T1) = -11 oC = 273-11 = 262 K

P1 = 6.65 torr

Now we will calculate the mole by applying the ideal gas equation:-

PV = nRT

Or, n = PV/RT

Where P is the pressure

T is the temperature

R is the gas constant = 0.0821 L atm mol-1K-1

V is the volume

Substituting the values of P, V, T, and R the mole of ethanol is calculated as:-

= 0.001913 mol C2H6

Conversion of the mole to gm

Molar mass of ethanol (M) = 46.07 g/mol

Mass of C2H6O =0.001913 mol C2H6O 46.07 g/mol = 0.088 = 8.8×10⁻²g.

Mass of ethanol in liquid form = total mass - a mass of ethanol in vapor

= 0.33g - 0.18113g

= 0.15 g (approx)

Hence, the mass of ethanol in the liquid phase is = 0.15 g

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