Question 27 (2 points) A student attempts to extract caffeine from tea bags by p

Question

Question 27 (2 points) A student attempts to extract caffeine from tea bags by partitioning between water and dichloromethane. The tea bags are boiled in water, removed and then the total volume of water made to be 50.0 mL. The partition coefficient for caffeine is 7.54 and the tea bags contained 53.1 mg of caffeine. If the student extracts the water once with 25.0 mL of dichloromethane, what mass of caffeine Will be in the dichloromethane? Report answer in mg to 3 sig figs. Question 28 (2 points) A binary mixture of 20 mol A and 88 mol B were heated to boiling. The equilibrium vapor pressures at boiling were 263 and 355, respectively. Determine the mole percentage of A in the vapor. Write answer in decimal form with three sig figs.

Explanation / Answer

Question 27

Assumed the partition coeff given is between DCM and water

Let x g of the organic compound is extracted by 25 ml DCM , leaving [ 53.1-x] mg in 50 ml water

So,[ x / 25 ] / [( 53.1 - x ) / 50] = 7.54

0.04 x / [ 1.062 - 0.02x ] = 7.54

0.04 x = 8.00748 -0.1508x

0.1908x = 8.00748

x = 41.97 mg

Answer: 41.97mg mass of caffeine will be there in DCM

Question 28.

mole fraction of A = 20/(20+88) = 0.185

mole fraction B = 88/(20+88) = 0.815

Using Raoult's Law,

Pa = XaPo (Xa is mole fraction of A , Po is the equilibrium vapour pressure)

For a Pa = 0.185 x 263 = 48.655

Pb = 0.815 x 355 = 289.325

P total = Pa + Pb = 48.655 + 289.325 = 337.98

In vapour phase,

Pa = y1 x Ptotal ( y1 is the mole fraction in vapour phase)

y1 = 48.655/337.98 =0.144

so mole fraction of a in vapour phase is 0.144

so, mole percentage of A in vapour is 14.39%

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