Question

20 g hemoglobin in a 1 L aqueous solution (A) at 300 K is separated from pure water by semi permeable membrane. At equilibrium the height of solution in a tube dipped in a solution (A) is found to be 80.0 mm higher than the tube dipped in water. The molar mass of hemoglobin is ______ kg mol$^{-1}$. (Nearest integer)
(Given : $g=10 \text{ m s}^{-2}$, $R=8.3 \text{ kPa dm}^3 \text{ K}^{-1}\text{mol}^{-1}$, density of solution $= 1000 \text{ kg m}^{-3}$)}

Answer 1

Correct Answer: 62

Step 1: Understanding the Concept:
Osmotic pressure ($\pi$) can be measured by the hydrostatic pressure exerted by the column of liquid at equilibrium, given by $\pi = h\rho g$. This pressure is also related to the concentration of the solute via the Van't Hoff equation $\pi = CRT$.
: Key Formula or Approach:
1. $\pi = h\rho g$.
2. $\pi = \frac{w}{MV} RT$.
Step 2: Detailed Explanation:
1. Calculate Osmotic Pressure ($\pi$):
Height $h = 80.0 \text{ mm} = 0.08 \text{ m}$.
Density $\rho = 1000 \text{ kg/m}^3$.
Gravity $g = 10 \text{ m/s}^2$.
$\pi = h\rho g = 0.08 \times 1000 \times 10 = 800 \text{ N/m}^2 = 800 \text{ Pa}$.
Convert to kPa: $\pi = 0.8 \text{ kPa}$.
2. Calculate Molar Mass (M):
Mass of solute $w = 20 \text{ g}$.
Volume $V = 1 \text{ L} = 1 \text{ dm}^3$.
Temperature $T = 300 \text{ K}$.
$R = 8.3 \text{ kPa dm}^3 \text{ K}^{-1} \text{mol}^{-1}$.
Using $\pi = \frac{w}{MV} RT$:
$0.8 = \frac{20}{M \cdot 1} \cdot 8.3 \cdot 300$.
$M = \frac{20 \cdot 8.3 \cdot 300}{0.8} = \frac{49800}{0.8} = 62250 \text{ g/mol}$.
3. Convert to kg/mol:
$M = 62.25 \text{ kg/mol}$.
Step 3: Final Answer:
The molar mass of hemoglobin is 62 kg mol$^{-1}$ to the nearest integer.