Answer all of the following questions. To receive full credit you must show all

Question

Answer all of the following questions. To receive full credit you must show all your work. Answers only will not receive credit for the problems. Pure water has freezing point : T-oc. A solution is prepared by dissolving 25.0 g sugar in 100 g of pure water. And this sugar solution has the freezing point: -2.6°C . what is the molar mass of the unknown sugar? (The freezing point depression constant : K,-1.86°C m-1 ) 1) 2. If 30 mL (30 g) of warm water (51.0 °C) are mixed with 30 mL (30 g) of cold water (20.2°C) in a calorimeter, the equilibrium temperature attained by the system is 34.8 The specific heat of water is 4.18J/g'C. Calculate the calorimeter constant for this system.

Explanation / Answer

Q1

Apply Colligative properties

This is a typical example of colligative properties.

Recall that a solute ( non volatile ) can make a depression/increase in the freezing/boiling point via:

dTf = -Kf*molality * i

dTb = Kb*molality * i

where:

Kf = freezing point constant for the SOLVENT; Kb = boiling point constant for the SOLVENT;

molality = moles of SOLUTE / kg of SOLVENT

i = vant hoff coefficient, typically the total ion/molecular concentration.

At the end:

Tf mix = Tf solvent - dTf

Tb mix = Tb solvent - dTb

substitute

mol of sugar = mass/MW = 25/MW=

kg of water = 100g = 0.1

molality =  25/MW/0.1 = 250/MW

Tf = -1.86 * 0.73099

-2.6= -1.86 *  250/MW

MW = 1.86/2.6*250 = 17884 g/mol

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