Question

Determine the vapour pressure (in atm) of the following solutions: a. 25.0g of ethyl alcohol (C2H5OH) in 100.0g of water at 25.0°C. Pvap (H2O, 25°C) = 23.8 mmHg; Pvap (C2H5OH, 25°C) = 61.2 mmHg b. 410g of sucrose (C12H22011) and 320g water at 25°C. Pvap (H20, 25°C) = 23.8 mmHg 5. The vapour pressures determined in question #4 are one of the four colligative properties of solutions. Determine the other three colligative properties of the following solution. Determine Th and T, in °C, and II in bar; Kr(H2O) = 1.858 °C kg/mol; K (H2O) = 0.5120 °C kg/mol): 7.400g of MgCl2 is dissolved in 110.0g of H20 at 25.0°C.

Answer 1

4 (a) Total vapour pressure of solution ,P = ( P_EtOH- P_H2O) $\chi$ _EtOH + P_H2O

$\Rightarrow$ P = (0.08 - 0.03) [0.54/(0.54+5.55)] + 0.03 = 0.034 atm ( $\chi$ _EtOH = mole fraction of ethanol = number of moles of ethanol/total number of moles of ethanol and water)

4(b) Total vapour pressure of solution ,P = ( P_sucrose- P_H2O) $\chi$ _sucrose + P_H2O

$\Rightarrow$ P = (0 - 0.03) [1.19/(1.19+17.77)] + 0.03 = 0.011 atm ( $\chi$ _EtOH = mole fraction of sucrose=number of moles of sucrose/total number of moles of sucrose and water)

5. MgCl₂ being a strong electolyte will completely ionize to form Mg²⁺ and 2Cl^- ions. Thus, i=3

Using the equation $\Delta$ T_b =iK_bm where K_b is the ebulioscopic constant and m is the molarity.

Substituting the values, we get $\Delta$ T_b =3*0.5120*(7.4/95.2) * (1000/110) =1.08

Thus, the boiling point of the solution is 100+1.08 = 101.08 degree celcius

Using the equation $\Delta$ T_f=iK_fm where K_f is the cryoscopic constant and m is the molarity.

Substituting the values, we get $\Delta$ T_f =3*1.858*(7.4/95.2) * (1000/110) =3.93

Thus, the freezing point of the solution is 100-3.93 = 96.07 degree celcius

Using the equation,$\pi$= iCRT =i(n/V)RT = {3 [(7.4/95.2)] *0.0821*298.15} / 0.11 = 51.40 atm = 52.08 bar

Thus, the osmotic pressure of the solution is 52.08 bar.