Question

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. (a) What mass of ethanol is present in the vapor?

Answer 1

The mass of ethanol present in the vapor is 8.8×10⁻²g. when liquid and vapor ethanol 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.

Hence, the mass of ethanol present in the vapor is found to be  8.8×10⁻²g.

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