What is the freezing point of a 0.27 m solution of glucose, C_6 H_12 O_6 in wate

Question

What is the freezing point of a 0.27 m solution of glucose, C_6 H_12 O_6 in water? (K_fp for water is 1.858 degree C/m.) What is the freezing point of a solution containing 2.80 grams benzene (molar mass = 78.11 g/mol) dissolved in 43.0 grams paradichlorobenzene (molar mass = 147.0 g/mol? The freezing point of pure paradichlorobenzene is 53.0 degree C and the freezing point depression constant, K_fp is -7.10 degree C/m. A solution consisting of 0.21 mol of methylbenzene, C_6 H_5 CH_3, in 259 g of nitrobenzene, C_6 H_5 NO_2, freezes at 0.3 degree C. Pure nitrobenzene freezes at 6.0 degree C. What is the freezing-point depression constant of nitrobenzene? A 2.5-g sample of a small protein having a molecular weight of 72,000 g/mol is dissolved in 32.0 mL of water at 25 degree C. What is the osmotic pressure of the solution? (R = 0.0821 L atm/K mol; 1 atm = 760 mmHg)

Explanation / Answer

Ans:

12)   T = imK ;i=van't Hoff factor; m=molality; K=1.858
T = (1)(0.24)(1.858) = 0.45 oC

new freezing point is -0.45 oC because freezing point of pure water is 0 oC.

13) Tf = Kf*m  ; Tf = change in freezing point

m = moles of solute / kg of solvent

moles of benzene = grams / Molecular mass = 2.8 / 78.11 = 0.0358 mol
m = 0.0358 / 0.043 = 0.8336 mol / kg
Tf = 0.8336 * 7.10 = 5.9 °C
Tf = 53 - 5.9 = 47.1 °C
So freezing point Tf = 47.1 °C

14)  molality = 0.210 mol / .259 kg = 0.8108 mole/ kg

Kf = t / m
= 5.7 / 0.8108 = 7.0 oC / molal.

15)  pi = i M R T

M = n/V;

n = 2.5/72000 = 3.472*10-5 mol

M = 3.472*10-5 mol / 0.032 L = .001085

pi = (1) (0.001085) (0.0821) (298) = 0.026 atm

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