What is the osmolarity of a solution that contains $4.00\%$ (m/v) $\ce{NaCl}$ $(M = \pu{58.44 g mol-1})$ and $3.00\%$ (m/v) glucose $(M = \pu{180.18 g mol-1})?$

I know you have to convert percentages to mass soultion/Liter solution and multiply by number of moles in $\ce{NaCl},$ which is 2 moles:

$\pu{2 osmol}$ $\ce{NaCl}$ / $\pu{1 mol}$ $\ce{NaCl}$

but I'm being thrown off by being given molar mass.

  • $\begingroup$ Hint: Osmolarity is osmotic molarity. Molarity is ... The adjective osmotic means... $\endgroup$ – Poutnik Oct 20 at 7:22

Let's begin by finding the molarity of each solute in that solution. We'll get to osmolarity later.


The concentration of NaCl given in the problem is $0.04 \frac{\text{g}}{\text{mL}}=40 \frac{\text{g}}{\text{L}}$. We can divide by the molar mass, getting $\frac{40\text{g}}{\text{L}}\cdot\frac{\text{mol}}{58.44\text{g}}\approx0.6845\text{ M.}$ (M represents molar, or mol/L.)


The concentration of glucose given in the problem is $0.03 \frac{\text{g}}{\text{mL}}=30 \frac{\text{g}}{\text{L}}$. We can divide by the molar mass, getting $\frac{30\text{g}}{\text{L}}\cdot\frac{\text{mol}}{180.18\text{g}}\approx0.1665\text{ M.}$

It is at this point that we consider the distinction between osmolarity and molarity.

According to Wikipedia,



  • $\phi$ is the osmotic coefficient, which accounts for the degree of non-ideality of >the >solution. In the simplest case it is the degree of dissociation of the solute. >Then, $\phi$ is between 0 and 1 where 1 indicates 100% dissociation. However, $\phi$ can >also be larger than 1 (e.g. for sucrose). For salts, electrostatic effects cause $\phi$ >to be smaller than 1 even if 100% dissociation occurs (see Debye–Hückel equation);
  • n is the number of particles (e.g. ions) into which a molecule dissociates.
  • C is the molar concentration of the solute;
  • the index i represents the identity of a particular solute.

For the moment, we are going to ignore $\phi$ and assume that that everything dissociates perfectly. We can make this assumption because glucose and NaCl generally dissolve near-completely in water.

From that, we get $\text{osmolarity}=\displaystyle\sum_{i}n_iC_i=n_\text{NaCl}C_\text{NaCl}+n_\text{glucose}C_\text{glucose}$

We know that NaCl dissociates into two ions: Na$^+$ and Cl$^-$, so $n_\text{NaCl}=2.$ Glucose, however, does not dissociate, but rather stays as a single molecule. Therefore, $n_\text{glucose}=1.$

We now have $\text{osmolarity}=2\cdot0.6845+1\cdot0.1665=\boxed{1.5355 \text{ osmolar}}$


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