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Solutions question

2008 · Shift 1 · Q21
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Solutions question

2008 · Shift 1 · Q21

JEE AdvancedChemistrySolutionsMCQ+3 / −1
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9. Given: Freezing point depression constant of water (Kfwater)=1.86\left( {K_f^{water}} \right) = 1.86(Kfwater​)=1.86 K kg mol −1{}^{-1}−1 Freezing point depression constant of ethanol (Kfethanol)=2.0\left( {K_f^{ethanol}} \right) = 2.0(Kfethanol​)=2.0 K kg mol −1{}^{-1}−1 Boiling point elevation constant of water (Kbwater)=0.52\left( {K_b^{water}} \right) = 0.52(Kbwater​)=0.52 K kg mol −1{}^{-1}−1 Boiling point elevation constant of ethanol (Kbethanol)=1.2\left( {K_b^{ethanol}} \right) = 1.2(Kbethanol​)=1.2 K kg mol −1{}^{-1}−1 Standard freezing point of water = 273 K Standard freezing point of ethanol = 155.7 K Standard boiling point of water = 373 K Standard boiling point of ethanol = 351.5 K Vapour pressure of pure water = 32.8 mm Hg Vapour pressure of pure water = 40 mm Hg Molecular weight of water = 18 g mol −1{}^{-1}−1 Molecular weight of ethanol = 46 g mol −1{}^{-1}−1 In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.The freezing point of the solution M is :
  1. A
    268.7 K
  2. B
    268.5 K
  3. C
    234.2 K
  4. D
    150.9 K
View written solutionFree

Correct answer: D

Step-by-step Solution:

1. Identify the Solvent and Solute

The solution M is a mixture of ethanol and water. The mole fraction of ethanol (XethanolX_{ethanol}Xethanol​) is given as 0.9. The sum of mole fractions in a binary mixture is 1. Therefore, the mole fraction of water (XwaterX_{water}Xwater​) is: Xwater=1−Xethanol=1−0.9=0.1X_{water} = 1 - X_{ethanol} = 1 - 0.9 = 0.1Xwater​=1−Xethanol​=1−0.9=0.1 Since the mole fraction of ethanol (0.9) is much larger than the mole fraction of water (0.1), ethanol is the solvent and water is the solute. The problem states to consider the solutions as ideal dilute solutions, which further supports that the component with the smaller mole fraction is the solute.

2. State the Formula for Freezing Point Depression

The depression in freezing point (ΔTf\\\Delta T_fΔTf​) is a colligative property and is calculated using the formula: ΔTf=Kf×m\Delta T_f = K_f \times mΔTf​=Kf​×m where:

  • KfK_fKf​ is the freezing point depression constant (cryoscopic constant) of the solvent.
  • mmm is the molality of the solute.

The freezing point of the solution (Tf,solutionT_{f, solution}Tf,solution​) is then found by: Tf,solution=Tf,pure solvent−ΔTfT_{f, solution} = T_{f, pure\ solvent} - \Delta T_fTf,solution​=Tf,pure solvent​−ΔTf​

3. Calculate the Molality (m) of the Solute

We need to find the molality of water (solute) in ethanol (solvent). Molality is defined as the number of moles of solute per kilogram of solvent. m=moles of solutemass of solvent (in kg)m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}m=mass of solvent (in kg)moles of solute​ Let's assume we have 1 mole of the solution in total.

  • Moles of ethanol (nethanoln_{ethanol}nethanol​) = Xethanol×1=0.9X_{ethanol} \times 1 = 0.9Xethanol​×1=0.9 mol
  • Moles of water (nwatern_{water}nwater​) = Xwater×1=0.1X_{water} \times 1 = 0.1Xwater​×1=0.1 mol

Now, we calculate the mass of the solvent (ethanol). The molar mass of ethanol (C2H5OHC_2H_5OHC2​H5​OH) is 46 g mol−1{}^{-1}−1. Mass of ethanol = nethanol×Molar mass of ethanoln_{ethanol} \times \text{Molar mass of ethanol}nethanol​×Molar mass of ethanol Mass of ethanol = 0.9 mol×46 g mol−1=41.40.9 \text{ mol} \times 46 \text{ g mol}^{-1} = 41.40.9 mol×46 g mol−1=41.4 g

Convert the mass of the solvent to kilograms: Mass of ethanol = 41.4 g=0.041441.4 \text{ g} = 0.041441.4 g=0.0414 kg

Now, calculate the molality (m): m=nwatermass of ethanol (kg)=0.1 mol0.0414 kg≈2.415 mol kg−1m = \frac{n_{water}}{\text{mass of ethanol (kg)}} = \frac{0.1 \text{ mol}}{0.0414 \text{ kg}} \approx 2.415 \text{ mol kg}^{-1}m=mass of ethanol (kg)nwater​​=0.0414 kg0.1 mol​≈2.415 mol kg−1

4. Calculate the Freezing Point Depression (ΔTf\\\Delta T_fΔTf​)

Using the formula for freezing point depression, with the values for the solvent (ethanol):

  • Kfethanol=2.0K_f^{ethanol} = 2.0Kfethanol​=2.0 K kg mol−1{}^{-1}−1
  • m=2.415m = 2.415m=2.415 mol kg−1{}^{-1}−1

ΔTf=Kfethanol×m=2.0 K kg mol−1×2.415 mol kg−1=4.83 K\Delta T_f = K_f^{ethanol} \times m = 2.0 \text{ K kg mol}^{-1} \times 2.415 \text{ mol kg}^{-1} = 4.83 \text{ K}ΔTf​=Kfethanol​×m=2.0 K kg mol−1×2.415 mol kg−1=4.83 K

5. Calculate the Freezing Point of the Solution

The standard freezing point of the pure solvent (ethanol) is given as Tf,pure ethanol=155.7T_{f, pure\ ethanol} = 155.7Tf,pure ethanol​=155.7 K.

Tf,solution=Tf,pure ethanol−ΔTfT_{f, solution} = T_{f, pure\ ethanol} - \Delta T_fTf,solution​=Tf,pure ethanol​−ΔTf​ Tf,solution=155.7 K−4.83 K=150.87 KT_{f, solution} = 155.7 \text{ K} - 4.83 \text{ K} = 150.87 \text{ K}Tf,solution​=155.7 K−4.83 K=150.87 K

6. Conclusion

The calculated freezing point of the solution M is 150.87 K, which is approximately 150.9 K. This matches option D.

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