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

2021 · 16 Mar · Shift 1 · Q20
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 Main
  3. /Chemistry
  4. /Solutions
  5. /2021 · 16 Mar · Shift 1 · Q20

Solutions question

2021 · 16 Mar · Shift 1 · Q20

JEE MainChemistrySolutionsNumerical+4 / −1
AB2AB_2AB2​ is 10% dissociated in water to A2+A^{2+}A2+ and B −-−. The boiling point of a 10.0 molal aqueous solution of AB2AB_2AB2​ is ‾\underline{\hspace{2cm}}​∘^\circ∘ C. (Round off to the Nearest Integer). [Given : Molal elevation constant of water Kb = 0.5 K kg mol −-− 1 boiling point of pure water = 100 ∘^\circ∘ C]
Numerical answer
View written solutionFree

Correct answer: 106

  1. Dissociation of the solute

Given:

  • Solute is AB2AB_2AB2​
  • Degree of dissociation α=10%=0.10\alpha = 10\% = 0.10α=10%=0.10

It dissociates as: AB2→A2++2B−AB_2 \rightarrow A^{2+} + 2B^-AB2​→A2++2B−

So, 1 formula unit of AB2AB_2AB2​ gives a total of 333 particles after complete dissociation.

  1. Van't Hoff factor

For dissociation into ν\nuν particles: i=1+(ν−1)αi = 1 + (\nu - 1)\alphai=1+(ν−1)α

Here, ν=3\nu = 3ν=3, so: i=1+(3−1)(0.10)=1+0.20=1.20i = 1 + (3-1)(0.10) = 1 + 0.20 = 1.20i=1+(3−1)(0.10)=1+0.20=1.20

  1. Boiling point elevation

Formula: ΔTb=iKbm\Delta T_b = i K_b mΔTb​=iKb​m

Given:

  • i=1.20i = 1.20i=1.20
  • Kb=0.5 K kg mol−1K_b = 0.5\ \text{K kg mol}^{-1}Kb​=0.5 K kg mol−1
  • m=10.0m = 10.0m=10.0

Therefore, ΔTb=1.20×0.5×10.0=6.0 ∘C\Delta T_b = 1.20 \times 0.5 \times 10.0 = 6.0\ ^\circ CΔTb​=1.20×0.5×10.0=6.0 ∘C

  1. Boiling point of solution

Boiling point of pure water = 100∘C100^\circ C100∘C

Hence, Tb(solution)=100+6=106∘CT_b(\text{solution}) = 100 + 6 = 106^\circ CTb​(solution)=100+6=106∘C

  1. Nearest integer

106\boxed{106}106​

PreviousNext

More from Solutions

  • At 363 K, the vapour pressure of A is 21 kPa and that of B is 18 kPa. One mole of A and 2 moles of B are mixed. Assuming that this solution is ideal, the vapour pressure of the mixture is ​ kPa. (Round off to the…2021 · Numerical
  • The oxygen dissolved in water exerts a partial pressure of 20 kPa in the vapour above water. The molar solubility of oxygen in water is ​× 10 − 5 mol dm − 3. (Round off to the Nearest Integer). [Given :…2021 · Numerical
  • A 1 molal K4​Fe(CN)6​ solution has a degree of dissociation of 0.4. Its boiling point is equal to that of another solution which contains 18.1 weight percent of a non electrolytic solute A. The molar mass of A is ​…2021 · Numerical
  • 2 molal solution of a weak acid HA has a freezing point of 3.885 ∘ C. The degree of dissociation of this acid is ​× 10 − 3. (Round off to the Nearest Integer). [Given : Molal depression constant of…2021 · Numerical
  • A solute A dimerizes in water. The boiling point of a 2 molal solution of A is 100.52 ∘ C. The percentage association of A is ​. (Round off to the Nearest Integer). [Use : Kb for water = 0.52 K kg mol − 1…2021 · Numerical
  • At 20 ∘ C, the vapour pressure of benzene is 70 torr and that of methyl benzene is 20 torr. The mole fraction of benzene in the vapour phase at 20 ∘ above an equimolar mixture of benzene and methyl benzene is ​…2021 · Numerical
  • The vapour pressures of A and B at 25 ∘ C are 90 mm Hg and 15 mm Hg respectively. If A and B are mixed such that the mole fraction of A in the mixture is 0.6, then the mole fraction of B in the vapour phase is x × 10 − 1.…2021 · Numerical
  • Which one of the following 0.06 M aqueous solutions has lowest freezing point?2021 · MCQ