a) Calculate the boiling point (in degrees C) of a solution made by dissolving 5

Question

a) Calculate the boiling point (in degrees C) of a solution made by dissolving 5.78 g of fructose (C6H12O6) in 12 g of water. The Kbp of the solvent is 0.512 K/m and the normal boiling point is 373 K. Enter your answer to 2 decimal places.

b) When 13.7 g of an unknown, non-volatile, non-electrolyte, X was dissolved in 100. g of methanol, the vapor pressure of the solvent decreased from 122.7 torr to 115 torr at 298 K. Calculate the molar mass of the solute, X

c) Determine the mass in grams of hexane that must be added to 61 g of benzene to make a 0.56 m solution. Give your answer to 2 decimal places.

Explanation / Answer

a) Molar mass of Fructose (C6H12O6) = 6 * 12 + 12 * 1 + 6 * 16 = 180 gm/mol

Number of moles of Fructose = 5.78/180 = 0.032111 moles

molality = moles of solute/solvent in Kg = (5.78/180)/0.012 = 2.675 m

New Boiling Point = Old boiling Point + i * Kb * m

=> 373K + 2.675 * 1 * 0.512

=> 374.37K

b)

Methanol(CH3OH) Molar mass = 12 + 4 * 1 + 16 = 32 gm/mol

Number of moles of methanol = 100/32 = 3.125 moles

Vapor pressure = Pure vapor pressure * mole fraction

mole fraction = 0.9372

Let the number of moles of solute be x

3.125/(3.125+x) = 0.9372

x = 0.209 moles

Molar mass of solute = mass/number of moles = 13.7/0.209 = 62.12 gm/mol

c) molaity = 0.56 m = number of moles of solute/(61/1000)

moles of solute = 0.03416 moles of Benzene

Molar mass of Benzen (C6H6) = 6 * 12 + 6 * 1 = 78 gm/mol

Mass of Benzene Required = 0.03416 moles * 78 gm/mol = 2.66 gms

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