6. Na2SO4 is an electrolyte. The dissolution of Na2SO4 in water can be described

Question

6. Na2SO4 is an electrolyte. The dissolution of Na2SO4 in water can be described by the chemical equation given below Na2SO4 (s)-> 2Na + (aq) + SO42-(aq) a) Solution A is ade by dissolving completely 10.0 grams of Na2SO4 in 80.0 mL of deionized water. The density of water is 1.00 g/mL and the freezing point of water is 0.0 °C. What is the expected freezing point of solution A? b) Solution B is made by dissolving completely 18.0 grams of Na-SO in 80.0 mL of deionized water. What is the expected freezing point of solution B? r, 'Allah Ana has a hinher freezinq point? Explain why.

Explanation / Answer

Q6)

a)

Molar mass of Na2SO4 = 23 * 2 + 32 + 4 * 16 = 142 gm/mol

Number of moles of Na2SO4 in 10 gms = Mass/molar mass = 10/142 = 0.07042 moles

Mass of solvent(water) = Volume of water * density of water = 80 mL * 1 gm/mL = 80 gms

molality of Na2SO4 = number of moles of solute/mass of solvent in Kg = (10/142)/(0.080) = 0.88028m

Depression in Freezing point = i * Kf * m

Kf = 1.86 C/m (constant value)

i value = 3 (since Na2SO4 dissociates into three ions,2 ions of Na+ and one ion of SO4(2-)]

Depression in Freezing Point = 3 * 1.86 * 0.88028 = 4.911C

Expected Freezing Point of solution = Actual Freezing Point - Depression in Freezing Point = 0 - 4.911 = -4.911C

b)

Molar mass of Na2SO4 = 23 * 2 + 32 + 4 * 16 = 142 gm/mol

Number of moles of Na2SO4 in 10 gms = Mass/molar mass = 18/142 = 0.12676 moles

Mass of solvent(water) = Volume of water * density of water = 80 mL * 1 gm/mL = 80 gms

molality of Na2SO4 = number of moles of solute/mass of solvent in Kg = (18/142)/(0.080) = 1.5845 m

Depression in Freezing point = i * Kf * m

Kf = 1.86 C/m (constant value)

i value = 3 (since Na2SO4 dissociates into three ions,2 ions of Na+ and one ion of SO4(2-)]

Depression in Freezing Point = 3 * 1.86 * 1.5845 = 8.841C

Expected Freezing Point of solution = Actual Freezing Point - Depression in Freezing Point = 0 - 8.841 = -8.841C

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