The temperature at which a solution freezes and boils depends on the freezing an

Question

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 boiling for a solution, delta T_b, can be calculated as delta T_b = K_b middot m in which m is the molality of the solution and K_b is the molal boiling-point-elevation constant for the solvent. The freezing-point depression, deltaT_f, can be calculated in a similar manner: delta T_f = 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 6.50 degree C and a K_f of 20.0 3 degree C/m. What is the freezing point of a solution made by dissolving 0.925 g of biphenyl (C_12H_10) in 25.0 g of cyclohexane? Express the temperature numerically in degrees Celsius. Paradichlorobenzene, C_6H_4C_12, 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. K_b =

Explanation / Answer

Part A) Depression in freezing point of solution dTf = iKfm

with,

i = 1

Kf = 20 oC/m

m = 0.925/154.21 x 0.025 = 0.24 m

we get,

dTf = 1 x 0.24 x 20 = 4.8 oC

Feezing point of solution = 6.5 - 4.8 = 1.7 oC

Part B) For elevation in boiling point,

dTb = iKbm

with,

i = 1

dTb = 82.39 - 80.70 = 1.69 oC

m = 2/147 x 0.0225 = 0.605 m

So,

Kb = 1.69/0.605 = 2.80 oC/m

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