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

2019 · Shift 1 · Q14
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. /2019 · Shift 1 · Q14

Solutions question

2019 · Shift 1 · Q14

JEE AdvancedChemistrySolutionsNumerical+3 / −1
On dissolving 0.5 g of a non-volatile non-ionic solute to 39 g of benzene, its vapor pressure decreases from 650 mmHg to 640 mmHg. The depression of freezing point of benzene (in K) upon addition of the solute is ............. (Given data: Molar mass and the molal freezing point depression constant of benzene are 78 g mol-1 and 5.12 K kg mol-1, respectively).
Numerical answer
View written solutionFree

Correct answer: 1.02

  1. Use relative lowering of vapour pressure

For a non-volatile solute, p0−pp0=x2\frac{p^0-p}{p^0}=x_2p0p0−p​=x2​ where x2x_2x2​ is the mole fraction of solute.

Given: p0=650 mmHg,p=640 mmHgp^0=650\text{ mmHg},\quad p=640\text{ mmHg}p0=650 mmHg,p=640 mmHg So, 650−640650=10650=165\frac{650-640}{650}=\frac{10}{650}=\frac{1}{65}650650−640​=65010​=651​ Hence, x2=165x_2=\frac{1}{65}x2​=651​

  1. Find moles of benzene

Mass of benzene =39 g=39\text{ g}=39 g

Molar mass of benzene =78 g mol−1=78\text{ g mol}^{-1}=78 g mol−1

n1=3978=0.5 moln_1=\frac{39}{78}=0.5\text{ mol}n1​=7839​=0.5 mol

  1. Find moles of solute

Using x2=n2n1+n2x_2=\frac{n_2}{n_1+n_2}x2​=n1​+n2​n2​​ we get n20.5+n2=165\frac{n_2}{0.5+n_2}=\frac{1}{65}0.5+n2​n2​​=651​

Solving, 65n2=0.5+n265n_2=0.5+n_265n2​=0.5+n2​ 64n2=0.564n_2=0.564n2​=0.5 n2=0.564=0.0078125 moln_2=\frac{0.5}{64}=0.0078125\text{ mol}n2​=640.5​=0.0078125 mol

  1. Calculate molality

Mass of solvent =39 g=0.039 kg=39\text{ g}=0.039\text{ kg}=39 g=0.039 kg

m=n2kg of solvent=0.00781250.039=0.20032 mol kg−1m=\frac{n_2}{\text{kg of solvent}}=\frac{0.0078125}{0.039}=0.20032\text{ mol kg}^{-1}m=kg of solventn2​​=0.0390.0078125​=0.20032 mol kg−1

  1. Use freezing point depression formula

ΔTf=Kfm\Delta T_f=K_f mΔTf​=Kf​m

Given Kf=5.12 K kg mol−1K_f=5.12\text{ K kg mol}^{-1}Kf​=5.12 K kg mol−1

ΔTf=5.12×0.20032≈1.026 K\Delta T_f=5.12\times 0.20032\approx 1.026\text{ K}ΔTf​=5.12×0.20032≈1.026 K

Therefore, ΔTf≈1.02 K\boxed{\Delta T_f\approx 1.02\text{ K}}ΔTf​≈1.02 K​

  1. Comparison with stored answer

Stored correct answer = 1.021.021.02

Our derived answer matches it.

PreviousNext

More from Solutions

  • Liquids A and B form ideal solution over the entire range of composition. At temperature T, equimolar binary solution of liquids A and B has vapor pressure 45Torr. At the same temperature, a new solution of A and B having mole…2018 · Numerical
  • The plot given below shows P−T curves (where P is the pressure and T is the temperature) for two solvents X and Y and isomolal solutions of NaCl in these solvents. NaCl completely dissociates in both the solvents. On addition… Includes diagram2018 · Numerical
  • For a solution formed by mixing liquids L and M, the vapor pressure of L plotted against the mole fraction of M in solution is shown in the following figure. Here XL​ and XM​ represent mole fractions of L and M,… Includes diagram2017 · Multiple correct
  • Pure water freezes at 273K and 1 bar. The addition of 34.5g of ethanol to 500g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2 kg mol−1. The figures shown…2017 · MCQ
  • The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is the same as its molality. Density of this solution at 298 K is 2.0 g cm–3 . The ratio of the molecular weights of the solute and solvent, (MWsolvent​MWsolute​​)…2016 · Numerical
  • The qualitative sketches I, II and III given below show the variation of surface tension with molar concentration of three different aqueous solutions of KCl, CH3​OH and CH3(CH2)11​ OSO 3−​ Na+ at room temperature. The… Includes diagram2016 · MCQ
  • Mixture (s) showing positive deviation from Raoult’s law at 35oC is (are)2016 · Multiple correct
  • If the freezing point of a 0.01 molal aqueous solution of a cobalt (III) chloride-ammonia complex(which behaves as a strong electrolyte) is – 0.0558oC, the number of chloride(s) in the coordination sphere of the complex is [Kf of water =…2015 · Numerical