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A glucose solution contains 55.8 g of glucose (c6h12o6) in 455 g of water. calculate the freezing point and boiling point of the solution. ( density of water = 1.00 g/ml, kb= 0.512 o c kg solvent/mol solute and 1.86°c kg/mol)

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Dejanras

Answer is: the freezing point of the solution of glucose is -1.26°C and boiling point is 100.353°C.
m(H₂O) = 455 g ÷ 1000 g/kg = 0.455 kg.
m(C₆H₁₂O₆) = 55.8 g. 

n(C₆H₁₂O₆) = m(C₆H₁₂O₆)÷ M(C₆H₁₂O₆).
n(C₆H₁₂O₆) = 55.8 g ÷ 180.16 g/mol.
n(C₆H₁₂O₆) = 0.31 mol.
b(solution) = n(C₆H₁₂O₆) ÷ m(H₂O).
b(solution) = 0.31 mol ÷ 0.455 kg.
b(solution) = 0.68 mol/kg.
ΔTf = b(solution) · Kf(H₂O).
ΔTf = 0.68 mol/kg · 1.86°C·kg/mol.

ΔTf = 1.26°C.
Tf = 0°C - 1.26°C = -1.26°C.

ΔTb = b(solution) · Kb(H₂O).

ΔTb = 0.68 mol/kg · 0.52°C·kg/mol.

ΔTb = 0.353°C.

Tb = 100°C + 0.353°C.

pstnonsonjoku

The boiling point of the solution is 100.35 °c.

We know that;

ΔT = K m i

ΔT = freezing point depression

K = freezing constant

m = molality

i = Van't Hoff factor

Hence;

ΔT =  1.86°c kg/mol × 55.8 g/180 g/mol × 1/0.455 × 1

ΔT = 1.27 °c

Freezing point = 0 - 1.27 °c = - 1.27 °c

For boiling point;

ΔT = K m i

ΔT = 0.512 o c kg × 55.8 g/180 g/mol × 1/0.455 × 1

ΔT = 0.35 °c

Boiling point = 100 +  0.35 °c = 100.35 °c

Learn more about freezing point: https://brainly.com/question/2292439

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