Question 13

NUMERICALMEDIUM

At 300 K300\text{ K}, an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in terms of the height (hh) of the solution (density =1.00 g cm−3= 1.00\text{ g cm}^{-3}) where hh is equal to 2.00 cm2.00\text{ cm}. If the concentration of the dilute solution of the macromolecule is 2.00 g dm−32.00\text{ g dm}^{-3}, the molar mass of the macromolecule is calculated to be X×104 g mol−1X \times 10^4\text{ g mol}^{-1}. The value of XX is ______.

Use: Universal gas constant (RR) =8.3 J K−1 mol−1= 8.3\text{ J K}^{-1}\text{ mol}^{-1} and acceleration due to gravity (gg) =10 m s−2= 10\text{ m s}^{-2}

Correct Answer: 2.49

Detailed Solution

  1. Calculate osmotic pressure (π\pi) using hydrostatic pressure formula: π=hρg\pi = h\rho g

h=2.00 cm=0.02 mh = 2.00\text{ cm} = 0.02\text{ m}

ρ=1.00 g cm−3=1000 kg m−3\rho = 1.00\text{ g cm}^{-3} = 1000\text{ kg m}^{-3}

g=10 m s−2g = 10\text{ m s}^{-2}

π=0.02×1000×10=200 Pa (or N m−2)\pi = 0.02 \times 1000 \times 10 = 200\text{ Pa (or N m}^{-2})

  1. Use the osmotic pressure formula π=CRT\pi = CRT where CC is molar concentration:

π=wMVRT\pi = \frac{w}{MV}RT

Given mass concentration wV=2.00 g dm−3=2.00 g/L=2000 g m−3\frac{w}{V} = 2.00\text{ g dm}^{-3} = 2.00\text{ g/L} = 2000\text{ g m}^{-3}

200=2000M×8.3×300200 = \frac{2000}{M} \times 8.3 \times 300

M=2000×8.3×300200M = \frac{2000 \times 8.3 \times 300}{200}

M=10×2490=24900 g mol−1M = 10 \times 2490 = 24900\text{ g mol}^{-1}

M=2.49×104 g mol−1M = 2.49 \times 10^4\text{ g mol}^{-1}

Therefore, X=2.49X = 2.49.

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