Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Solutions question

2008 · Shift 1 · Q23
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Advanced
  3. /Chemistry
  4. /Solutions
  5. /2008 · Shift 1 · Q23

Solutions question

2008 · Shift 1 · Q23

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.Water is added to the solution M such that the fraction of water in the solution becomes 0.9 mole. The boiling point of this solution is:
  1. A
    380.4 K
  2. B
    376.2 K
  3. C
    375.5 K
  4. D
    354.7 K
View written solutionFree

Correct answer: B

Step-by-step Solution:

1. Determine the composition of the new solution.

The problem starts with a solution M (0.9 mole fraction ethanol, 0.1 mole fraction water). Then, water is added to this solution to create a new solution where the mole fraction of water is 0.9.

For this new solution:

  • Solvent: Water (since its mole fraction is higher)
  • Solute: Ethanol
  • Mole fraction of water (xwaterx_{water}xwater​) = 0.9
  • Mole fraction of ethanol (xethanolx_{ethanol}xethanol​) = 1 - xwaterx_{water}xwater​ = 1 - 0.9 = 0.1

2. Calculate the molality (m) of the solution.

Molality is defined as the number of moles of solute per kilogram of solvent. We can calculate it from the mole fractions.

Let's assume we have a total of 1 mole of the solution. Then:

  • Moles of solute (ethanol), nethanoln_{ethanol}nethanol​ = 0.1 mol
  • Moles of solvent (water), nwatern_{water}nwater​ = 0.9 mol

Now, we need the mass of the solvent (water) in kilograms.

  • Molar mass of water (MwaterM_{water}Mwater​) = 18 g mol−1^{-1}−1
  • Mass of water = nwater×Mwatern_{water} \times M_{water}nwater​×Mwater​ = 0.9 mol ×\times× 18 g mol−1^{-1}−1 = 16.2 g
  • Mass of water in kg = 16.2 g / 1000 g/kg = 0.0162 kg

Now, we can calculate the molality (m): m=moles of solutemass of solvent in kg=nethanolmass of water in kgm = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{n_{ethanol}}{\text{mass of water in kg}}m=mass of solvent in kgmoles of solute​=mass of water in kgnethanol​​ m=0.1 mol0.0162 kg≈6.173 mol kg−1m = \frac{0.1 \text{ mol}}{0.0162 \text{ kg}} \approx 6.173 \text{ mol kg}^{-1}m=0.0162 kg0.1 mol​≈6.173 mol kg−1

3. Calculate the boiling point elevation (\\[\Delta T_b\\]).

The elevation in boiling point is given by the formula: ΔTb=Kb×m\Delta T_b = K_b \times mΔTb​=Kb​×m Here, the solvent is water, so we use the boiling point elevation constant for water, KbwaterK_b^{water}Kbwater​.

  • Given: Kbwater=0.52K_b^{water} = 0.52Kbwater​=0.52 K kg mol−1^{-1}−1

ΔTb=0.52 K kg mol−1×6.173 mol kg−1\Delta T_b = 0.52 \text{ K kg mol}^{-1} \times 6.173 \text{ mol kg}^{-1}ΔTb​=0.52 K kg mol−1×6.173 mol kg−1 ΔTb≈3.21 K\Delta T_b \approx 3.21 \text{ K}ΔTb​≈3.21 K

4. Calculate the new boiling point of the solution (Tb′T_b'Tb′​).

The boiling point of the solution is the boiling point of the pure solvent plus the elevation.

  • Standard boiling point of water (TbwaterT_b^{water}Tbwater​) = 373 K

Tb′=Tbwater+ΔTbT_b' = T_b^{water} + \Delta T_bTb′​=Tbwater​+ΔTb​ Tb′=373 K+3.21 KT_b' = 373 \text{ K} + 3.21 \text{ K}Tb′​=373 K+3.21 K Tb′=376.21 KT_b' = 376.21 \text{ K}Tb′​=376.21 K

5. Compare with the options.

The calculated boiling point is 376.21 K, which matches option B (376.2 K).

Note: The problem asks to treat the solute (ethanol) as non-volatile, which simplifies the calculation to the standard colligative property formula, even though ethanol is actually volatile.

PreviousNext

More from Solutions

  • When 20 g of naphthoic acid (C 11​ H 8​ O 2​) is dissolved in 50 g of benzene in 50 g of benzene (K f​ = 1.72 K kg mol −1), a freezing point depression of 2 K is observed. The van't Hoff factor (i) is :2007 · MCQ
  • At 300 K , an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in terms of the height (h) of the solution (density =1.00 g cm−3) where h is equal to 2.00 cm . If the…2025 · Numerical
  • Vessel-1 contains w2​ g of a non-volatile solute X dissolved in w1​ g of water. Vessel- 2 contains w2​ g of another non-volatile solute Y dissolved in w1​ g…2024 · Numerical
  • 50 mL of 0.2 molal urea solution (density =1.012 g mL−1 at 300 K) is mixed with 250 mL of a solution containing 0.06 g of urea. Both the solutions were prepared in…2023 · Numerical
  • An aqueous solution is prepared by dissolving 0.1 mol of an ionic salt in 1.8 kg of water at 35∘C. The salt remains 90% dissociated in the solution. The vapour pressure of the solution is 59.724 mm…2022 · Numerical
  • The boiling point of water in a 0.1 molal silver nitrate solution (solution A) is x ∘ C. To this solution A, an equal volume of 0.1 molal aqueous barium chloride solution is added to make a new solution B. The difference in the…2021 · Numerical
  • The boiling point of water in a 0.1 molal silver nitrate solution (solution A) is x ∘ C. To this solution A, an equal volume of 0.1 molal aqueous barium chloride solution is added to make a new solution B. The difference in the…2021 · Numerical
  • Liquids A and B form ideal solution for all compositions of A and B at 25 ∘ C. Two such solutions with 0.25 and 0.50 mole fractions of A have the total vapour pressure of 0.3 and 0.4 bar, respectively. What is the vapour pressure…2020 · Numerical