Glycerol (C_3H_8O_3), also called glycerine, is widely used in the food and pharmaceutical industries. Glycerol is polar and dissolves readily in water and polar organic solvents like ethanol. Calculate the mole fraction of the solvent in a solution that contains 1.31 g glycerol dissolved in 24.10 mL ethanol (CH_3CH_2OH: density = 0.7893 g/mL). Round to four significant digits. X_solvent = 0.9437 Use your answer to Part 2 to calculate the vapor pressure of the solution at 20 degree C if the vapor pressure of pure ethanol at this temperature is equal to 0.0575 atm. Assume that glycerol is a nonvolatile, nonelectrolyte solute. Round your answer to three significant digits. P_solution = 0.0543 atm

1. 0.972.

2. 0.0559 atm.

Explanation:

1.

Solution of glycerol and ethanol.

Mole fraction of solvent = mole of solvent/total moles conposed in the solution

Total moles compsed in the solution = moles of glycerol + moles of ethanol

Number of moles:

Ethanol

Molar mass of ethanol = (12*2) + (1*6) + 16

= 46 g/mol.

Mass of ethanol = density * volume

= 0.7893 * 24.10

= 19.022 g

Moles = mass/molar mass

= 19.022/46

= 0.4135 mol.

Glycerol

Molar mass = (12*3) + (1*8) + (16*3)

= 92 g/mol

Moles = mass/molar mass

= 1.31/92

= 0.012 mol.

Total moles in the solution = 0.012 + 0.4135

= 0.4255 mol.

Mole fraction of ethanol(solvent) = 0.4135/0.4255

= 0.972

2.

Vapour pressure of ethanol, Peth = 0.0575 atm

Vapour pressure of solution, Psol = Xsolv * P°solv

Where,

P°solv = vapour pressure of the solvent, ethanol

Xsolv = mole fraction of solvent, ethanol

Po is the vapour pressure of the pure solvent at a particular temperature.

= 0.972 * 0.0575

= 0.0559 atm

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole