2. The boling pointis 78 26 C and the molal bolling point constant is 1.22 Cmola

Question

2. The boling pointis 78 26 C and the molal bolling point constant is 1.22 Cmolal for pure ethanol. (10 points each) A) if a solution with 2.455 g of an unknown solute(non-electrolyte) dissolved in 100.0 g of ethanol(CHCHOH) has a boiling point of 78.50, what is the molar mass of the unknown solute? 8) 13.50 grams of potassium sulfate is dissolved in 100.0 g of ethanol what would be the treezing point of the solution?-11 4.50C and 0.76 /molal is the freezing point and molal freezing point constant for ethanol.

Explanation / Answer

Ans 2

A)

Elevation in boiling point =kb. m

Elevation = 78.50 - 78.26 = 0.24oC

0.24 = 1.22 x m

m = 0.197 m

Molality = no. of moles / massof solvent in kg

0.197 = n / 0.100

n = 0.0197 moles

Mass of 0.0197 moles is 2.455 g

So the molar mass = 2.455 / 0.0197

= 124.62 g/mol

B)

number of moles of potassium sulfate = 3.50 / 174.259 = 0.02 moles

Molality = 0.02 / 0.100

m = 0.2 m

Depression in freezing point = i.kf. m

= 3 x0.76 x 0.2

= 0.456oC

So the freezing point = -114.50oC -0.456oC

= -114.956oC

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