Question

0.78 g of sodium hydroxide (NaOH) are added to a 1 L solution of water at a temperature of 21°C in a tube separated by a semi-permeable membrane. Calculate the osmotic pressure of the solution.

Answer 1

The osmotic pressure of a solution can be calculated according to the equation:

$\Pi = icRT$ ,

using $i$ - van't Hoff factor, $c$ - the molar concentration of the solute, $R$ - the ideal gas constant(0.082057 L·atm/mol·K), $T$ - the temperature of the solution in kelvins.

Let's convert the temperature first:

$T = 21($°C$) + 273.15 = 294.15$ K.

The molar concentration of the solute (NaOH, molar mass is 40.00 g/mol) is:

$c = \frac{n}{V} = \frac{m}{M·V} = \frac{0.78(\text{g})}{40.00\text{(g/mol)}·1\text{(L)}} = 0.0195 \text{(mol/L)}$.

The van't Hoff factor of strong electrolytes is equal to the number of ions produced from the dissociation of one formula unit. From NaOH formula unit, 2 ions are produced: Na⁺ and OH^-. Therefore, the van't Hoff factor of NaOH is equal to 2.

Finally, the osmotic pressure is:

$\Pi = 2·0.0195 \text{(mol/L)}·0.082057 \text{(L atm/(mol K))}·294.15 \text{(K)} = 0.94 \text{ atm}$

Answer: the osmotic pressure of the solution is 0.94 atm.