Question

Osmosis is a process by which the molecules of a solvent pass from a solution of low solute concentration to a solution of high solute concentration through a semi-permeable membrane. Osmotic pressure is a colligative property. When the applied pressure on a solution exceeds its osmotic pressure, reverse osmosis occurs. When two solutions are separated by a semipermeable membrane and they have same osmotic pressure they are said to be isotonic. Of the two solutions separated by a semipermeable membrane, if one is a lower osmotic pressure, it is said to be hypotonic relative to the second solution. If it has a higher osmotic pressure, than the second solution, it is said to be hypertonic relative to the second solution. The osmotic pressure associated with the fluid inside the blood cell is equivalent to that of 0.9 solution called normal saline solution and it is safe to inject intravenously. Osmotic pressure is vital in daily life and nature. It helps in explain, why IV fluids match blood’s osmotic pressure, its also the principle behind food preservation using salt or sugar. Calculate the amount of $CaCl_2$ ($i = 2.59$) dissolved in 2.46 litre of water such that its osmotic pressure is 0.70 atm at $27^\circ C$.
$[$Given: $R = 0.082~L~atm~K^{-1}mol^{-1}$, molar mass of $CaCl_2 = 111~g~mol^{-1}]$

Hint:

Always convert Temperature to Kelvin ($^\circ C + 273$) before using the ideal gas constant $R$.

Answer 1

Step 1: Osmotic pressure formula with van't Hoff factor.
For an electrolyte: $\pi = iCRT$. Since $C = \frac{w}{MV}$ (molarity), this gives $\pi = \frac{i \cdot w \cdot R \cdot T}{M \cdot V}$. Rearranging for mass: $w = \frac{\pi \cdot M \cdot V}{i \cdot R \cdot T}$.
Step 2: List given values.
$\pi = 0.70~atm$, $M = 111~g~mol^{-1}$, $V = 2.46~L$, $i = 2.59$, $R = 0.082~L~atm~K^{-1}~mol^{-1}$, $T = 27 + 273 = 300~K$.
Step 3: Calculate numerator and denominator.
Numerator: $0.70 \times 111 \times 2.46 = 191.142$. Denominator: $2.59 \times 0.082 \times 300 = 63.714$.
Step 4: Calculate mass.
\[ w = \frac{191.142}{63.714} \approx 3.0~g \] The mass of $CaCl_2$ required is $3.0~g$.