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Solutions question

2025 · 3 Apr · Shift 1 · Q10
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Solutions question

2025 · 3 Apr · Shift 1 · Q10

JEE MainChemistrySolutionsMCQ+4 / −1
2 moles each of ethylene glycol and glucose are dissolved in 500 g of water. The boiling point of the resulting solution is: (Given : Ebullioscopic constant of water =0.52 K kg mol−1=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}=0.52 K kg mol−1 )
  1. A
    377.3 K
  2. B
    379.2 K
  3. C
    375.3 K
  4. D
    277.3 K
View written solutionFree

Correct answer: A

  1. Use elevation in boiling point formula

For a non-volatile solute,

ΔTb=Kb m\Delta T_b = K_b\, mΔTb​=Kb​m

where mmm is the total molality of all solute particles.

  1. Find total moles of solute

Given:

  • Ethylene glycol =2= 2=2 moles
  • Glucose =2= 2=2 moles

Both are non-electrolytes, so they do not dissociate.

Thus, total moles of solute:

2+2=4 mol2+2=4\text{ mol}2+2=4 mol
  1. Mass of solvent in kg

Water =500 g=0.5 kg= 500\text{ g} = 0.5\text{ kg}=500 g=0.5 kg

  1. Calculate molality
m=moles of solutekg of solvent=40.5=8 mol kg−1m = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{4}{0.5} = 8\text{ mol kg}^{-1}m=kg of solventmoles of solute​=0.54​=8 mol kg−1
  1. Calculate elevation in boiling point

Given:

Kb=0.52 K kg mol−1K_b = 0.52\, \text{K kg mol}^{-1}Kb​=0.52K kg mol−1

So,

ΔTb=0.52×8=4.16 K\Delta T_b = 0.52 \times 8 = 4.16\text{ K}ΔTb​=0.52×8=4.16 K
  1. Boiling point of solution

Normal boiling point of water:

373 K373\text{ K}373 K

Hence,

Tb=373+4.16=377.16 KT_b = 373 + 4.16 = 377.16\text{ K}Tb​=373+4.16=377.16 K

This is closest to:

377.3 K377.3\text{ K}377.3 K
  1. Check options
  • A: 377.3 K377.3\text{ K}377.3 K ✅
  • B: 379.2 K379.2\text{ K}379.2 K
  • C: 375.3 K375.3\text{ K}375.3 K
  • D: 277.3 K277.3\text{ K}277.3 K

Therefore, the correct option is A.

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