19: Suppose 2.000 kg of sugar is dissolved in enough water to make S density of

Question

19: Suppose 2.000 kg of sugar is dissolved in enough water to make S density of the solution is 1.123 g/mL. Find: A: Calculate the molality and the molarity of the solution. B: Find the Pop lowering and the Ppop of the new solvent (After mixing): Given the Pwp of pure water if 23.0 torr at this temperature. c: Find the boiling point elevation and boiling point of water after mixing. Given: the normal BP of water is 100.00 C D: Find the freezing point depression and the freezing point after mixing: Given the FP of water is 0.00 C

Explanation / Answer

Density of solution = 1.123 g/mL

Volume of solution = 5000 mL = 5 L

Mass of solution = 1.123 * 5000 = 5615 g = 5.615 kg

Mass of sugar dissolved = 2 kg = 2000 g

Mass of water in solution = 5.615 – 2 kg = 3.615 kg

Moles of water in solution = 3615 g / 18 g/mol = 200.8 moles

Sugar is C12H22O11

Molar mass of sugar = 342.3 g/mol

Moles of sugar dissolved = 2000 g / 342.3 g/mol = 5.84

A)

Molarity of sugar, M = Moles of sugar / Volume of solution

= 5.84 / 5 L = 1.17 M

Molality of sugar, m = Moles of sugar / Mass of water (kg)

= 5.84 / 3.615 = 1.62 m

B)

For non-volatile solute,

Pvap = Povap X

X = mole fraction of solvent (water)

Pvap = 23 torr * (200.8 / (200.8 + 5.84))

= 22.35 torr

Pvap lowering = 23 – 22.35 torr = 0.65 torr

C)

For non-ionic compounds, the vant hoff factor, i = 1

Tb = i Kb m

For water Kb = 0.512 °C · kg H2O / mol solute

Boiling point elevation, Tb = 1 * 0.512 * 1.62 = 0.83 °C

Boiling point of water, Tb = 100 + Tb

= 100 + 0.83 = 100.83 °C

D)

For non-ionic compounds, the vant hoff factor = 1

Tf = i Kf m

For water, Kf = 1.86 °C/m

Freezing point depression, Tf = 1 * 1.86 * 1.62 = 3 °C

Freezing point of water, Tf = 0 - Tf

= 0 - 3 = - 3 °C

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