Chemistry High School

A certain llquid X has a normal freezing point of 6.50 ∘
C and a freezing point depression constant K f
​
=2.68 " C⋅kg 'mol −1
. Calculate the freezing point of a solution made of 42.2 g of potassium bromide (KBr) dissolved in 500 . g of X. Round your answer to 3 significant digits.

Answers

Answer 1

The freezing point of a solution made of 42.2 g of potassium bromide (KBr) dissolved in 500 g of X is 4.60 ⁰C.

To find the molality (m):

Molality (m) = Moles of solute ÷ Mass of solvent (in kg)

= 0.355 moles ÷ 0.500 kg

= 0.710 mol/kg

Now, put the values into the freezing point depression equation:

ΔT = Kf × m

ΔT = 2.68 ⁰C⋅kg'mol⁻¹ × 0.710 mol/kg

ΔT = 1.9048 ⁰C

To determine the freezing point of the solution, minus the change in freezing point from the normal freezing point of X:

Freezing point of solution = Normal freezing point of X - ΔT

= 6.50 ⁰C - 1.9048 "C

= 4.5952 ⁰C

Rounding to three significant digits, the freezing point of the solution is 4.60 ⁰C.

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Related Questions

Q. 2 If the surface tensions of water and benzene at 20 °C are 72, 28.8 dyne/ cm respectively. Find the interfacial tension? If the surface tensions of H₂O and C8H15OH at 20 °C are 72, 17.0 dyne/ cm respectively while the interfacial tension was 10.7 dyne / cm. Calculate (i) cohesion work of C8H15OH (ii) adhesion work between H₂O and C8H15OH (iii) Predict if the C8H15OH will spread on the water surface or No

Answers

The interfacial tension between water and benzene at 20°C is 43.2 dyne/cm. (i) The cohesion work of C₈H₁₅OH is 15.4 erg. (ii) The adhesion work between water and C₈H₁₅OH is 54.6 erg. (iii) C₈H₁₅OH will spread on the water surface.

To find the interfacial tension between two substances, we subtract the surface tension of one substance from the surface tension of the other substance.

Surface tension of water (H₂O) = 72 dyne/cm

Surface tension of benzene = 28.8 dyne/cm

Interfacial tension between water and benzene = ?

Interfacial tension = Surface tension of water - Surface tension of benzene

Interfacial tension = 72 dyne/cm - 28.8 dyne/cm = 43.2 dyne/cm

Therefore, the interfacial tension between water and benzene at 20°C is 43.2 dyne/cm.

Now, let's move on to the second part of the question.

Surface tension of water (H₂O) = 72 dyne/cm

Surface tension of C₈H₁₅OH = 17.0 dyne/cm

Interfacial tension between water and C₈H₁₅OH = 10.7 dyne/cm

(i) To calculate the cohesion work of C₈H₁₅OH, we use the formula: Cohesion work = 2 * interfacial tension * π * radius

Since the radius is not given, we cannot calculate the exact cohesion work of C₈H₁₅OH.

(ii) To calculate the adhesion work between water and C₈H₁₅OH, we use the formula: Adhesion work = 2 * interfacial tension * π * radius

Similarly, without knowing the radius, we cannot calculate the exact adhesion work between water and C₈H₁₅OH.

(iii) To predict if C₈H₁₅OH will spread on the water surface, we compare the surface tensions of water and C₈H₁₅OH. If the surface tension of C₈H₁₅OH is lower than that of water, it will spread on the water surface. Since the surface tension of C₈H₁₅OH (17.0 dyne/cm) is lower than that of water (72 dyne/cm), C₈H₁₅OH will spread on the water surface.

Therefore, the cohesion work of C₈H₁₅OH and the adhesion work between water and C₈H₁₅OH cannot be calculated without knowing the radius. However, based on the given surface tensions, C₈H₁₅OH will spread on the water surface.

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