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 · 26 Aug · Shift 2 · 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 · 26 Aug · Shift 2 · Q20

Solutions question

2021 · 26 Aug · Shift 2 · Q20

JEE MainChemistrySolutionsNumerical+4 / −1
83 g of ethylene glycol dissolved in 625 g of water. The freezing point of the solution is ‾\underline{\hspace{2cm}}​ K. (Nearest integer) [Use : Molal Freezing point depression constant of water = 1.86 K kg mol −-− 1] Freezing Point of water = 273 K Atomic masses : C : 12.0 u, O : 16.0 u, H : 1.0 u]
Numerical answer
View written solutionFree

Correct answer: 269

  1. Use freezing point depression formula

For a non-electrolyte solute: ΔTf=Kf⋅m\Delta T_f = K_f \cdot mΔTf​=Kf​⋅m where:

  • Kf=1.86 K kg mol−1K_f = 1.86\ \text{K kg mol}^{-1}Kf​=1.86 K kg mol−1
  • mmm = molality of solution
  1. Find molar mass of ethylene glycol

Ethylene glycol has formula C2H6O2\mathrm{C_2H_6O_2}C2​H6​O2​.

So, molar mass: M=2(12)+6(1)+2(16)=24+6+32=62 g mol−1M = 2(12) + 6(1) + 2(16) = 24 + 6 + 32 = 62\ \text{g mol}^{-1}M=2(12)+6(1)+2(16)=24+6+32=62 g mol−1

  1. Calculate moles of ethylene glycol

Given mass of ethylene glycol = 838383 g

moles=8362=1.3387 mol\text{moles} = \frac{83}{62} = 1.3387\ \text{mol}moles=6283​=1.3387 mol

  1. Calculate mass of solvent in kg

Mass of water = 625625625 g = 0.6250.6250.625 kg

  1. Calculate molality

m=1.33870.625=2.1419 mol kg−1m = \frac{1.3387}{0.625} = 2.1419\ \text{mol kg}^{-1}m=0.6251.3387​=2.1419 mol kg−1

  1. Calculate depression in freezing point

ΔTf=1.86×2.1419=3.984 K\Delta T_f = 1.86 \times 2.1419 = 3.984\ \text{K}ΔTf​=1.86×2.1419=3.984 K

  1. Find freezing point of solution

Freezing point of pure water = 273273273 K

Tf(solution)=273−3.984=269.016 KT_f(\text{solution}) = 273 - 3.984 = 269.016\ \text{K}Tf​(solution)=273−3.984=269.016 K

Nearest integer: 269\boxed{269}269​

  1. Comparison with stored answer

Stored correct answer = 269269269

This matches the derived answer.

PreviousNext

More from Solutions

  • 224 mL of SO2​(g) at 298 K and 1 atm is passed through 100 mL of 0.1 M NaOH solution. The non-volatile solute produced is dissolved in 36g of water. The lowering of vapour pressure of solution (assuming the solution in dilute) (P (H2​O)o​−…2021 · Numerical
  • When 12.2 g of benzoic acid is dissolved in 100 g of water, the freezing point of solution was found to be − 0.93 ∘ C (Kf(H2​O) = 1.86 K kg mol − 1). The number (n) of benzoic acid molecules associated (assuming 100%…2021 · Numerical
  • 1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to − 4 ∘ C before freezing. The amount of ice (in g) that will be separated out is ​. (Nearest integer) [Given : Kf(H2​O) = 1.86 K kg mol −…2021 · Numerical
  • 40 g of glucose (Molar mass = 180) is mixed with 200 mL of water. The freezing point of solution is ​ K. (Nearest integer) [Given : Kf = 1.86 K kg mol − 1; Density of water = 1.00 g cm − 3; Freezing point of…2021 · Numerical
  • 1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of 2.42 × 10 − 3 bar. The molar mass of the biopolymer is ​× 104 g mol − 1. (Round off to the Nearest Integer)…2021 · Numerical
  • In a solvent 50% of an acid HA dimerizes and the rest dissociates. The van't Hoff factor of the acid is ​× 10 − 2. (Round off to the nearest integer)2021 · Numerical
  • Which one of the following 0.10 M aqueous solutions will exhibit the largest freezing point depression?2021 · MCQ
  • 1.22 g of an organic acid is separately dissolved in 100 g of benzene (Kb = 2.6 K kg mol − 1) and 100 g of acetone (Kb = 1.7 K kg mol − 1). The acid is known to dimerize in benzene but remain as a monomer in acetone. The boiling point…2021 · Numerical