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Solutions question

2025 · Shift 2 · Q13
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Solutions question

2025 · Shift 2 · Q13

JEE AdvancedChemistrySolutionsNumerical+4 / −1
At 300 K , an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in terms of the height (h)(h)(h) of the solution (density =1.00 g cm−3=1.00 \mathrm{~g} \mathrm{~cm}^{-3}=1.00 g cm−3) where hhh is equal to 2.00 cm . If the concentration of the dilute solution of the macromolecule is 2.00 gdm−32.00 \mathrm{~g} \mathrm{dm}^{-3}2.00 gdm−3, the molar mass of the macromolecule is calculated to be X×104 g mol−1\boldsymbol{X} \times 10^4 \mathrm{~g} \mathrm{~mol}^{-1}X×104 g mol−1. The value of X\boldsymbol{X}X is ‾\underline{\hspace{2cm}}​. Use: Universal gas constant (R)=8.3 J K−1 mol−1(R)=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}(R)=8.3 J K−1 mol−1 and acceleration due to gravity (g)=10 m s−2(g)=10 \mathrm{~m} \mathrm{~s}^{-2}(g)=10 m s−2
Numerical answer
View written solutionFree

Correct answer: 2.4TO2.55

  1. Relate osmotic pressure to hydrostatic pressure

Given that osmotic pressure is expressed as the height hhh of the solution column,

π=ρgh\pi = \rho g hπ=ρgh

where:

  • ρ=1.00 g cm−3=1000 kg m−3\rho = 1.00\ \text{g cm}^{-3} = 1000\ \text{kg m}^{-3}ρ=1.00 g cm−3=1000 kg m−3
  • g=10 m s−2g = 10\ \text{m s}^{-2}g=10 m s−2
  • h=2.00 cm=2.00×10−2 mh = 2.00\ \text{cm} = 2.00 \times 10^{-2}\ \text{m}h=2.00 cm=2.00×10−2 m

So,

π=1000×10×2.00×10−2=200 Pa\pi = 1000 \times 10 \times 2.00 \times 10^{-2} = 200\ \text{Pa}π=1000×10×2.00×10−2=200 Pa

  1. Use the osmotic pressure formula

For a dilute solution,

π=cRT\pi = cRTπ=cRT

where ccc is molar concentration in mol m−3\text{mol m}^{-3}mol m−3.

Also,

c=mass concentrationMc = \frac{\text{mass concentration}}{M}c=Mmass concentration​

Given mass concentration:

2.00 g dm−3=2.00 kg m−32.00\ \text{g dm}^{-3} = 2.00\ \text{kg m}^{-3}2.00 g dm−3=2.00 kg m−3

Thus,

c=2.00Mc = \frac{2.00}{M}c=M2.00​

with MMM in kg mol−1\text{kg mol}^{-1}kg mol−1.

Hence,

π=2.00MRT\pi = \frac{2.00}{M}RTπ=M2.00​RT

Substitute values:

200=2.00M×8.3×300200 = \frac{2.00}{M} \times 8.3 \times 300200=M2.00​×8.3×300

200=4980M200 = \frac{4980}{M}200=M4980​

M=4980200=24.9 kg mol−1M = \frac{4980}{200} = 24.9\ \text{kg mol}^{-1}M=2004980​=24.9 kg mol−1

  1. Convert to g mol−1\text{g mol}^{-1}g mol−1

24.9 kg mol−1=24.9×103 g mol−1=2.49×104 g mol−124.9\ \text{kg mol}^{-1} = 24.9 \times 10^3\ \text{g mol}^{-1} = 2.49 \times 10^4\ \text{g mol}^{-1}24.9 kg mol−1=24.9×103 g mol−1=2.49×104 g mol−1

So,

X=2.49X = 2.49X=2.49

  1. Final answer

2.49\boxed{2.49}2.49​

Next

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