Osmotic pressure of a solution is 12 atm. What is the concentration of NaCl solution which is isotonic to the given solution at 900 K.

R = 0.082 Lit-atm K K^–1 mol^–1

Assume 100% dissociation.

`) are strictly followed.0.02439M

The correct option is 0.02439M.

Step 1: Understand the Condition for Isotonic Solutions

Two solutions are said to be isotonic when they exert the same osmotic pressure ($\pi$) at a given temperature. Therefore, the osmotic pressure of the NaCl solution must equal the given osmotic pressure of 12 atm.

Step 2: Formula for Osmotic Pressure

The osmotic pressure of a solution containing an electrolyte is given by the formula:

$\pi =\; i\; ·\; C\; ·\; R\; ·\; T$

Where:

• $\pi$ = Osmotic pressure = 12 atm
• $i$ = van 't Hoff factor
• $C$ = Molar concentration (molarity) of the solution
• $R$ = Solution gas constant = 0.082 Lit-atm K^–1 mol^–1
• $T$ = Absolute temperature = 900 K

Step 3: Determine the van 't Hoff Factor (i) for NaCl

NaCl undergoes 100% dissociation in aqueous solution into two ions:

$NaCl\; \Rightarrow \; Na++\; Cl–$

Since 1 mole of NaCl produces 2 moles of ions, the van 't Hoff factor $i\; =\; 2$.

Step 4: Calculate the Concentration (C)

Rearranging the osmotic pressure formula to solve for concentration $C$:

$C\; =\frac{\pi }{}$

Substituting the given values into the equation:

$C\; =\frac{12}{2\times 0.082\times 900\times 3.333}$

$C\; =\frac{12}{492}$

$C\; =\; 0.02439\; M$

Thus, the required concentration of the isotonic NaCl solution is 0.02439M.