Boiling Point Elevation and Freezing Point Depression for Organic Solutions The

Question

Boiling Point Elevation and Freezing Point Depression for Organic Solutions The temperature at which a solution freezes and boils depends on the freezing and boiling points of the pure solvent as well as on the molal concentration of particles (molecules and ions) in the solution. For nonvolatile solutes the boiling point of the solution is higher than that of the pure solvent and the freezing point is lower. The change in the baling fee a solution, Delta T_b, can be calculated as Delta T_b = K_b middot m in which m is the molaity of the solution and K_b is the molal boiling-point-elevation constant for the solvent. The freezing-point depression, Delta T_f, can be calculated in a similar manner Delta T_t = K_f middot m in which m is the molality of the solution and K_f is the molal freezing-point-depression constant for the solvent. Cyclohexane has a freezing point of 6.60 degree C and a K_f of 20.0 degree C/m. What is the freezing point of a solution made by dissolving 0.540 r of biphenyl? (C_12 H_10) in 25.0 g of cyclohexane? Express the temperature numerically in degrees Celsius. Paradichlorobenzene, C_6H_4C1_2 is a component of mothballs. A solution of 2.00 g in 22.5 g of cyclohexane boils at 82.39 degree C. The boiling point of pure cyclohexane is 80.70 degree C. Calculate K_b for cyclohexane. Express the constant numerically in degrees Celsius per molal.

Explanation / Answer

we know that

moles = mass / molar mass

so

moles of C12H10 = 0.54 / 154 = 3.5065 x 10-3

now

molality = moles x 1000 (g) / mass of solvent (g)

so

molality (m) = 3.5065 x 10-3 x 1000 / 25

molality (m) = 0.14

now

dTf = Kf x m

dTf = 20 x 0.14

dTf = 2.8

6.5 - T = 2.8

T = 3.7

so

freezing point of the solution is 3.7 C

B)

elevation in boiling point is given by

dTb = 82.39 - 80.70

dTb = 1.69

now

moles of C6H4Cl2 = 2 / 147 = 0.0136

now

molality (m) = moles x 1000 / mass of solvent (g)

so

molality (m) = 0.0136 x 1000 / 22.5

molality (m)= 0.60468

now

dTb = Kb x m

1.69 = Kb x 0.60468

Kb = 2.795

so

the value of Kb is 2.795

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