Q.$1.0\ \text{g}$ of a non-volatile, non-electrolyte solute of molar mass $60\ \text{g mol}^{-1}$ is dissolved in enough water to make $200\ \text{mL}$ of solution. Calculate the osmotic pressure of the solution at $300\ \text{K}$. ($R = 0.0821\ \text{L atm mol}^{-1}\text{K}^{-1}$.)

When a solution is separated from its pure solvent by a semipermeable membrane (permeable to solvent, not to

solute), solvent flows spontaneously into the solution side — osmosis — because part of the membrane adjacent

to the solution is effectively shielded by non-crossing solute particles, so relatively fewer solvent molecules

strike it from that side. The minimum external pressure that must be applied to the solution to stop this net

inflow exactly is the osmotic pressure, $\pi$, given for a dilute solution by van't Hoff's equation, which

takes the same form as the ideal gas law:

Rearranged, $M_2 = w_2 RT/(\pi V)$. Two solutions of equal $\pi$ are isotonic; a lower-$\pi$ solution is

hypotonic and a higher-$\pi$ solution is hypertonic relative to a reference — red blood cells swell (and

can burst) in a strongly hypotonic solution and shrink in a strongly hypertonic one, which is why intravenous …