Question
The osmotic pressure of a living cell is $12\text{ atm}$ at $300\text{ K}$. The strength of sodium chloride solution that is isotonic with the living cell at this temperature is __________ $\text{g L}^{-1}$. (Nearest integer)
Given : $\text{R} = 0.08\text{ L atm K}^{-1}\text{ mol}^{-1}$
Assume complete dissociation of $\text{NaCl}$
(Given : Molar mass of $\text{Na}$ and $\text{Cl}$ are $23$ and $35.5\text{ g mol}^{-1}$ respectively.)
Osmotic Pressure Calculation for Isotonic Solution
The problem requires finding the concentration of a sodium chloride ($\text{NaCl}$) solution that is isotonic with a living cell, meaning they have the same osmotic pressure at a given temperature.
We can use the van't Hoff equation for osmotic pressure:
$ \Pi = iCRT $
Where:
- $ \Pi $ is the osmotic pressure ($12 \text{ atm}$).
- $ i $ is the van't Hoff factor. For $\text{NaCl}$, which dissociates completely into $\text{Na}^{+}$ and $\text{Cl}^{-}$, $i = 2$.
- $ C $ is the molar concentration (in $\text{mol L}^{-1}$).
- $ R $ is the ideal gas constant ($0.08 \text{ L atm K}^{-1}\text{ mol}^{-1}$).
- $ T $ is the absolute temperature ($300 \text{ K}$).
Calculating Molar Concentration
Rearranging the formula to solve for $ C $:
$ C = \frac{\Pi}{iRT} $
Substituting the given values:
$ C = \frac{12 \text{ atm}}{(2) \times (0.08 \text{ L atm K}^{-1}\text{ mol}^{-1}) \times (300 \text{ K})} $
$ C = \frac{12}{0.16 \times 300} \text{ mol L}^{-1} $
$ C = \frac{12}{48} \text{ mol L}^{-1} $
$ C = 0.25 \text{ mol L}^{-1} $
Converting Molar Concentration to Mass Concentration
First, calculate the molar mass of $\text{NaCl}$:
- Molar mass of $\text{Na} = 23 \text{ g mol}^{-1}$
- Molar mass of $\text{Cl} = 35.5 \text{ g mol}^{-1}$
- Molar mass of $\text{NaCl} = 23 + 35.5 = 58.5 \text{ g mol}^{-1}$
Now, convert the molar concentration ($C$) to mass concentration (in $\text{g L}^{-1}$) using the molar mass ($M_{\text{NaCl}}$):
Mass Concentration = $ C \times M_{\text{NaCl}} $
Mass Concentration = $ 0.25 \text{ mol L}^{-1} \times 58.5 \text{ g mol}^{-1} $
Mass Concentration = $ 14.625 \text{ g L}^{-1} $
Final Answer
The question asks for the answer as the nearest integer.
Rounding $14.625 \text{ g L}^{-1}$ to the nearest integer gives $15 \text{ g L}^{-1}$.
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