JEE Main 2026 — Chemistry Physical Chemistry

The osmotic pressure $\pi$π of the solution is balanced by the hydrostatic pressure of the liquid column.

$$\pi = \rho g h$$π=ρgh

Given:

Density of solution, $\rho = 1000 \text{ kg m}^{-3}$ρ=1000 kg m−3

Acceleration due to gravity, $g = 10 \text{ m s}^{-2}$g=10 m s−2

Height difference, $h = 80.0 \text{ mm} = 0.08 \text{ m}$h=80.0 mm=0.08 m

$$\pi = 1000 \times 10 \times 0.08 = 800 \text{ Pa} = 0.8 \text{ kPa}$$π=1000×10×0.08=800 Pa=0.8 kPa

From the van 't Hoff equation for osmotic pressure:

$$\pi = C R T = \dfrac{n}{V} R T$$π=CRT=Vn​RT

Given:

Volume of solution, $V = 1 \text{ L} = 1 \text{ dm}^3$V=1 L=1 dm3

Gas constant, $R = 8.3 \text{ kPa dm}^3 \text{ K}^{-1} \text{mol}^{-1}$R=8.3 kPa dm3 K−1mol−1

Temperature, $T = 300 \text{ K}$T=300 K

Substituting the values:

$$0.8 = \dfrac{n}{1} \times 8.3 \times 300$$0.8=1n​×8.3×300

$$n = \dfrac{0.8}{2490} \text{ mol}$$n=24900.8​ mol

The mass of hemoglobin is $w = 20 \text{ g} = 0.02 \text{ kg}$w=20 g=0.02 kg.

The molar mass $M$M in $\text{kg mol}^{-1}$kg mol−1 is:

$$M = \dfrac{w}{n} = \dfrac{0.02}{\dfrac{0.8}{2490}}$$M=nw​=24900.8​0.02​

$$M = \dfrac{0.02 \times 2490}{0.8} = \dfrac{49.8}{0.8} = 62.25 \text{ kg mol}^{-1}$$M=0.80.02×2490​=0.849.8​=62.25 kg mol−1

Rounding to the nearest integer, we get $62$62.

Answer: $62$62