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NVAEasyJEE 2023Colligative Properties

JEE Chemistry 2023 Question with Solution

An aqueous solution of volume 300 cm3300 \, \text{cm}^3 contains 0.63 g0.63 \, \text{g} of protein. The osmotic pressure of the solution at 300 K300 \, \text{K} is 1.29 mbar1.29 \, \text{mbar}. The molar mass of the protein is _____ g mol−1\text{g mol}^{-1}.

Given: R=0.083 L bar K−1 mol−1R = 0.083 \, \text{L bar K}^{-1} \text{ mol}^{-1}

Answer

Correct answer:40535

Step-by-step solution

Standard Method

Given:

  • Volume V=300 cm3=0.3 LV = 300 \, \text{cm}^3 = 0.3 \, \text{L}
  • Mass m=0.63 gm = 0.63 \, \text{g}
  • Osmotic pressure π=1.29 mbar=1.29×10−3 bar\pi = 1.29 \, \text{mbar} = 1.29 \times 10^{-3} \, \text{bar}
  • Temperature T=300 KT = 300 \, \text{K}
  • Gas constant R=0.083 L bar K−1 mol−1R = 0.083 \, \text{L bar K}^{-1} \text{ mol}^{-1}

Find: The molar mass of the protein.

Using the formula:

π=cRT\pi = cRT

where cc is the molarity.

Calculate molarity:

c=πRT=1.29×10−3 bar0.083 L bar K−1mol−1×300 Kc = \frac{\pi}{RT} = \frac{1.29 \times 10^{-3} \, \text{bar}}{0.083 \, \text{L bar K}^{-1} \text{mol}^{-1} \times 300 \, \text{K}} c≈5.18×10−5 mol/Lc \approx 5.18 \times 10^{-5} \, \text{mol/L}

Calculate moles:

n=c×V=5.18×10−5 mol/L×0.3 Ln = c \times V = 5.18 \times 10^{-5} \, \text{mol/L} \times 0.3 \, \text{L} n≈1.554×10−5 moln \approx 1.554 \times 10^{-5} \, \text{mol}

Calculate molar mass:

M=mn=0.63 g1.554×10−5 molM = \frac{m}{n} = \frac{0.63 \, \text{g}}{1.554 \times 10^{-5} \, \text{mol}} M≈40540 g/molM \approx 40540 \, \text{g/mol}

Therefore, the molar mass of the protein is approximately 4054040540. The solution lists the correct answer as 4053540535.

Using osmotic pressure to find molar mass

Given: osmotic pressure data and solution volume are provided.

Find: molar mass of the protein.

First find the molarity from osmotic pressure relation:

π=cRT\pi = cRT

So,

c=πRTc = \frac{\pi}{RT}

Substitute the values:

c=1.29×10−30.083×300c = \frac{1.29 \times 10^{-3}}{0.083 \times 300} c≈5.18×10−5 mol/Lc \approx 5.18 \times 10^{-5} \, \text{mol/L}

Now use volume V=0.3 LV = 0.3 \, \text{L} to get number of moles:

n=cVn = cV n=5.18×10−5×0.3n = 5.18 \times 10^{-5} \times 0.3 n≈1.554×10−5 moln \approx 1.554 \times 10^{-5} \, \text{mol}

Now compute molar mass:

M=mnM = \frac{m}{n} M=0.631.554×10−5M = \frac{0.63}{1.554 \times 10^{-5}} M≈4.054×104 g mol−1M \approx 4.054 \times 10^4 \, \text{g mol}^{-1}

Thus the molar mass is about 4.05×1044.05 \times 10^4 and the accepted numerical answer is 4053540535.

Common mistakes

  • Using 300 cm3300 \, \text{cm}^3 directly as liters is incorrect because the gas constant is given in liters. Convert first: 300 cm3=0.3 L300 \, \text{cm}^3 = 0.3 \, \text{L}.

  • Using 1.29 mbar1.29 \, \text{mbar} as 1.29 bar1.29 \, \text{bar} is incorrect because mbar must be converted to bar. Use 1.29 mbar=1.29×10−3 bar1.29 \, \text{mbar} = 1.29 \times 10^{-3} \, \text{bar}.

  • Taking osmotic pressure formula as π=RTc\pi = \frac{RT}{c} is incorrect. For dilute solutions, the correct relation is π=cRT\pi = cRT, so first find cc and then the moles.

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