Question

The osmotic pressure of a living cell is 12 atm at 300 K. The strength of sodium chloride solution that is isotonic with the living cell at this temperature is ____________ g L$^{-1}$. (Nearest integer)
Given: R = 0.08 L atm K$^{-1}$ mol$^{-1}$
Assume complete dissociation of NaCl
(Given : Molar mass of Na and Cl are 23 and 35.5 g mol$^{-1}$ respectively.)

Hint:

Isotonic means $\pi_{1} = \pi_{2}$. Always check if the solute is an electrolyte to include the van't Hoff factor $i$. For calculations involving $R=0.0821$, using $1/12$ can often simplify fractions, but here $R=0.08$ was specifically given.

Answer 1

Correct Answer: 15

Step 1: Understanding the Concept:
Isotonic solutions possess the same osmotic pressure at a given temperature.
The osmotic pressure ($\pi$) of an electrolyte solution is given by $\pi = iCRT$.

Step 2: Key Formula or Approach:
1. $\pi = iCRT$.
2. Strength (g/L) = Molarity ($C$) $\times$ Molar mass ($M$).
3. For NaCl, $i = 2$ (since it dissociates into $Na^{+}$ and $Cl^{-}$).
4. Molar mass of NaCl = $23 + 35.5 = 58.5$ g/mol.

Step 3: Detailed Explanation:
Given $\pi = 12$ atm, $T = 300$ K, and $R = 0.08$ L atm K$^{-1}$ mol$^{-1}$.
Equating the osmotic pressure of the cell to the NaCl solution:
\[ 12 = 2 \times C \times 0.08 \times 300 \]
\[ 12 = C \times 48 \]
\[ C = \frac{12}{48} = 0.25 \text{ mol L}^{-1} \]
Now, calculating the strength:
\[ \text{Strength} = 0.25 \text{ mol L}^{-1} \times 58.5 \text{ g mol}^{-1} \]
\[ \text{Strength} = 14.625 \text{ g L}^{-1} \]

Step 4: Final Answer:
Rounding 14.625 to the nearest integer, we get 15.