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Solutions question

2019 · 12 Apr · Shift 2 · Q9
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Solutions question

2019 · 12 Apr · Shift 2 · Q9

JEE MainChemistrySolutionsMCQ+4 / −1
A solution is prepared by dissolving 0.6 g of urea (molar mass = 60 g mol–1) and 1.8 g of glucose (molar mass = 180 g mol–1) in 100 mL of water at 27oC. The osmotic pressure of the solution is : (R = 0.08206 L atm K–1 mol–1)
  1. A
    8.2 atm
  2. B
    2.46 atm
  3. C
    4.92 atm
  4. D
    1.64 atm
View written solutionFree

Correct answer: C

  1. Use osmotic pressure formula

For a dilute solution, π=CRT\pi = CRTπ=CRT where:

  • CCC = total molarity of solute particles
  • R=0.08206 L atm K−1 mol−1R = 0.08206\, \text{L atm K}^{-1}\text{ mol}^{-1}R=0.08206L atm K−1 mol−1
  • T=27∘C=300 KT = 27^\circ\text{C} = 300\,\text{K}T=27∘C=300K

Since urea and glucose are non-electrolytes, they do not dissociate. So total concentration is based on total moles dissolved.

  1. Calculate moles of urea

Mass of urea = 0.6 g0.6\,\text{g}0.6g

Molar mass of urea = 60 g mol−160\,\text{g mol}^{-1}60g mol−1

nurea=0.660=0.01 moln_{\text{urea}} = \frac{0.6}{60} = 0.01\,\text{mol}nurea​=600.6​=0.01mol

  1. Calculate moles of glucose

Mass of glucose = 1.8 g1.8\,\text{g}1.8g

Molar mass of glucose = 180 g mol−1180\,\text{g mol}^{-1}180g mol−1

nglucose=1.8180=0.01 moln_{\text{glucose}} = \frac{1.8}{180} = 0.01\,\text{mol}nglucose​=1801.8​=0.01mol

  1. Total moles of solute

ntotal=0.01+0.01=0.02 moln_{\text{total}} = 0.01 + 0.01 = 0.02\,\text{mol}ntotal​=0.01+0.01=0.02mol

  1. Volume of solution

Given 100 mL100\,\text{mL}100mL water. For this level of problem, we take solution volume as approximately V=100 mL=0.1 LV = 100\,\text{mL} = 0.1\,\text{L}V=100mL=0.1L

  1. Calculate molarity

C=nV=0.020.1=0.2 mol L−1C = \frac{n}{V} = \frac{0.02}{0.1} = 0.2\,\text{mol L}^{-1}C=Vn​=0.10.02​=0.2mol L−1

  1. Calculate osmotic pressure

π=CRT=(0.2)(0.08206)(300)\pi = CRT = (0.2)(0.08206)(300)π=CRT=(0.2)(0.08206)(300)

π=4.9236 atm\pi = 4.9236\,\text{atm}π=4.9236atm

∴π≈4.92 atm\therefore \pi \approx 4.92\,\text{atm}∴π≈4.92atm

  1. Check options
  • A: 8.2 atm8.2\,\text{atm}8.2atm ❌
  • B: 2.46 atm2.46\,\text{atm}2.46atm ❌
  • C: 4.92 atm4.92\,\text{atm}4.92atm ✅
  • D: 1.64 atm1.64\,\text{atm}1.64atm ❌

Hence, the correct option is C.

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