# Osmolarity and osmotic pressure

You have discovered a new strain of bacteria called Tankeromyces subtilis that has a special outer membrane that is capable of withstanding high levels of osmotic pressure. These bacteria can survive osmotic pressure as high $$\pu{1.3 kJ L-1}$$, but not higher. The $$(iC)_\mathrm{total}$$ value for the inside of these bacteria is $$\pu{630 mM}$$.

What is the lowest concentration of $$\ce{NaCl}$$ (in otherwise pure water) in which these bacteria can survive without bursting due to osmotic pressure?

$$R = \pu{8.3153 J mol-1 K-1}$$ and $$T = \pu{296 K}$$.

The answer is $$\pu{0.10 M}$$, but I have no idea how to drive at that answer. However, I do know that it involves this equation:

$$p = iCRT$$

• The units of osmotic pressure given may be confusing you. kJ/L is a pressure unit becaues a Joule is equal to a Pascal-cubic meter. $\textrm{J}=\textrm{Pa} \cdot \textrm{m}^3$ Does that give you enough to get started? – Curt F. Apr 22 '15 at 3:33
• Ohhh. I didn't even think of it that way. Yes that was very helpful in getting started. – Michael Apr 22 '15 at 11:35

The highest pressure they can handle is $\pu{1.3 kJ L-1}$, or $\pu{13000 J L-1}$. In the formula you use the ideal gas constant $0.08206$ and the temperature given to find the concentration the maximum pressure it can handle.
This comes out at $\pu{0.53 M}$ or $\pu{530 mM}$ and that's the maximum concentration it can handle. To find the minimum, you subtract $530$ from $630$ giving you $\pu{100 mM}$ or $\pu{0.1 M}$, which is the lowest concentration it can handle.