Question

1 litre of urea solution (1M) is mixed with 2 litre of 0.3M NaCl solution both at TK, osmotic pressure would be-

• A. 0.73 RT
• B. 0.8 RT
• C. 0.65 RT
• D. 0.5 RT

Answer 1

Answer: (A)

0.73 RT

Answer 2

The osmotic pressure ($\pi$) of a solution is given by the formula $\pi = iCRT$, where $i$ is the van't Hoff factor, $C$ is the molar concentration of the solute, $R$ is the ideal gas constant, and $T$ is the temperature in Kelvin.

When two solutions are mixed, the total osmotic pressure is the sum of the osmotic pressures contributed by each solute, assuming ideal mixing and ideal solution behavior. $\pi_{\text{total}} = \sum \pi_j = \sum i_j C_j RT$

First, calculate the number of moles of each solute in the initial solutions. Moles of urea = Volume × Molarity = $1 \text{ L} \times 1 \text{ M} = 1 \text{ mol}$. Moles of NaCl = Volume × Molarity = $2 \text{ L} \times 0.3 \text{ M} = 0.6 \text{ mol}$.

Next, calculate the total volume of the mixture. Total volume $V_{\text{total}} = 1 \text{ L} + 2 \text{ L} = 3 \text{ L}$.

Now, calculate the molar concentration of each solute in the final mixture. Concentration of urea $C_{\text{urea}} = \frac{\text{Moles of urea}}{\text{Total volume}} = \frac{1 \text{ mol}}{3 \text{ L}} = \frac{1}{3} \text{ M}$. Concentration of NaCl $C_{\text{NaCl}} = \frac{\text{Moles of NaCl}}{\text{Total volume}} = \frac{0.6 \text{ mol}}{3 \text{ L}} = 0.2 \text{ M}$.

Determine the van't Hoff factor ($i$) for each solute. Urea is a non-electrolyte, so $i_{\text{urea}} = 1$. NaCl is a strong electrolyte that dissociates into two ions (Na⁺ and Cl⁻) in water. Assuming complete dissociation, $i_{\text{NaCl}} = 2$.

Calculate the effective concentration ($iC$) for each solute in the mixture. Effective concentration of urea = $i_{\text{urea}} \times C_{\text{urea}} = 1 \times \frac{1}{3} \text{ M} = \frac{1}{3} \text{ M}$. Effective concentration of NaCl = $i_{\text{NaCl}} \times C_{\text{NaCl}} = 2 \times 0.2 \text{ M} = 0.4 \text{ M}$.

The total effective concentration of solute particles in the mixture is the sum of the effective concentrations of individual solutes. $C_{\text{effective, total}} = C_{\text{effective, urea}} + C_{\text{effective, NaCl}} = \frac{1}{3} \text{ M} + 0.4 \text{ M}$. To add these values, convert 0.4 to a fraction: $0.4 = \frac{4}{10} = \frac{2}{5}$. $C_{\text{effective, total}} = \frac{1}{3} + \frac{2}{5} = \frac{1 \times 5}{3 \times 5} + \frac{2 \times 3}{5 \times 3} = \frac{5}{15} + \frac{6}{15} = \frac{5+6}{15} = \frac{11}{15} \text{ M}$.

Finally, calculate the total osmotic pressure using the formula $\pi_{\text{total}} = C_{\text{effective, total}} RT$. $\pi_{\text{total}} = \frac{11}{15} RT$.

To compare this value with the given options, calculate the decimal value of $\frac{11}{15}$. $\frac{11}{15} \approx 0.7333...$

So, the osmotic pressure is approximately $0.7333 RT$.

The calculated value $0.7333 RT$ is closest to $0.73 RT$.