WORKSHOP 13 COLLIGATIVE PROPERTIES Show calculation setups and answers for all p

Question

WORKSHOP 13 COLLIGATIVE PROPERTIES Show calculation setups and answers for all problems below. 1. List the following aqueous solutions in the order of expected DECREASING FREEZING POINT: 0.075 m glucose; 0.075 m LiBr:0.030 m Zn(NO,) The normal freezing point of pure naphthalene is measured to be 80.29 when 3221 g of the nonelectrolyte urea (CH,N,O) is dissolved in 751.36 g of naphthalene, the freezing point is measured to be 75.34°C, what is the molal freezing point depression constant (K) for naphthalene? 2. when 132.0 g of CH(P-33.96 torr) and 147.0 g of GH,CL (P. 224.9 torr) are combined, what is the total vapor pressure of the ideal solution? 3, Workshop 131 Colligative Properties

Explanation / Answer

1) To calculate depression in freezing point, formula used is:

Tfo -Tf = iKfm

where Tfo = Freezing point of water

Tf = Freezing point of solution

i = vant Hoff factor

Kf = cryoscopic constant for water

m = molality of solution

Freezing point of water and cryoscopic constant is same for all the solutions.

For 0.075 m glucose, i = 1, Kf = 1.86 oC kg/mol, m = 0.075, Tfo = 0 oC

0-Tf = 1*1.86*0.075

Tf = -0.14 oC

Similarly, for 0.075 m LiBr, i = 2, Kf = 1.86 oC kg/mol, m = 0.075, Tfo = 0 oC

0-Tf = 2*1.86*0.075

Tf = -0.28 oC

For 0.030 m Zn(Br)2, i = 3, Kf = 1.86 oC kg/mol, m = 0.030, Tfo = 0 oC

0-Tf = 3*1.86*0.030

Tf = -0.17 oC

So, Decreasing freezing point is:

0.075 m LiBr > 0.030 m Zn(Br)2 > 0.075 m glucose

2) Tfo = 80.29 oC

Tf = 75.34 oC

i = 1 (for urea)

Kf = to be calculated

m = molality

Moles of urea = mass of urea/molar mass of urea = 32.21/60 = 0.637 mol

mass of solvent = mass of naphthalene = 751.36 g = 0.75136 kg

Molality = moles of urea/mass of solvent = 0.637/0.75136 = 0.85 mol/kg

So,

Tfo -Tf = iKfm

80.29-75.34 = 1*Kf*0.85

Kf = 5.82 oC kg/mol

Molal freezing point depression constant = 5.82 oC kg/mol

3) Mass of C6H6 = 132.0 g

molar mass of C6H6 = 78 g/mol

Moles of C6H6 = n1 = mass/molar mass = 132.0/78 = 1.69 mol

Mass of C2H4Cl2 = 147.0 g

molar mass of C2H4Cl2 = 99 g/mol

Moles of C2H4Cl2 = n2 = mass/molar mass = 147.0/99 = 1.48 mol

mole fraction of C6H6 = x1 = n1/(n1+n2) = 1.69/(1.69+1.48) = 0.53

So, x2 = 1-0.53 = 0.47

Vapour pressure of solution = x1P1 +x2P2

= (0.53*93.96)+(0.47*224.9) = 155.50 torr

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