Calculate the freezing point of a 14 m aqueous solution of ethylene glycol. The

Question

Calculate the freezing point of a 14 m aqueous solution of ethylene glycol. The k_f for water is 1.86 K kg/mol. -52 degree C -26 degree C -76 degree C -0.11 degree C 0 degree C In order to calculate the molarity of 1.00 M glucose(aq), the molar mass of water is required. volume of solution must be known. mole fraction of water must be given. mole fraction of glucose must be given. density of the solution must be known. A 1.0 m aqueous solution of an unknown solute has a freezing point of -9.30 degree C. Which of the following is the solute? The k_1 for water is 1.86 K kg/mol. calcium chloride iron (III) chloride sodium sulfate aluminum sulfate sodium chloride Calculate the boiling point elevation of an aqueous solution which is 2.20 m calcium per chlorate. The k_b for water is 0.510 K kg/mol. 0.560 degree C 0.510 degree C 1.12 degree C 2.24 degree C 3.37 degree C Blood, sweat and tears are about 0.15 M in sodium chloride. Estimate the osmotic pressure of these solutions at 37 degree C. The gas constant is 0.08206 L atm/K mol. 11 atm 7.6 atm 3.8 atm 1.8 atm 0.91 atm

Explanation / Answer

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29)

Depression in Freezing Point = i(vont hoff factor) * Kf * m(molality)

=> 1(for ethylene glycol) * 1.86 K.kg/mol ) 14 mol/Kg

=> 26

Hence the freezing point = 0 - 26 = -26C

Hence the correct answer is Option B

30)

The density of the solution must be known in order to calculate molality from molarity

Hence the correct answer is Option E

31)

Depression in Freezing Point = i(vont hoff factor) * Kf * m(molality)

9.30 = i * 1.86 * 1

i = 5

The vont hoff factor of 5 implies the compound dissociates into 5 ions, hence the correct answer is Option D

Al2(SO4)3 ---> 2Al(3+) + 3SO4(2-)

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