Question

What is the osmolarity of .00001 grams (0.1 mg%) of ethanol (does not dissociate) in one liter?

Answer 1

Explanation:

As it is known that non-electrolytes do not dissociate. Therefore, molarity of such a solution is equal to the osmolarity of solution.

As, molar mass of ethanol = 46.07 g/mol

Therefore, no. of moles of ethanol will be calculated as follows.

               No. of moles = $\frac{mass}{\text{molar mass}}$

                                     = $\frac{0.00001 g}{46.07 g/mol}$

                                     = $2.17 \times 10^{-7}$ mol

As, molarity is moles of solute in liter of solution. Hence, molarity of ethanol is as follows.

                           Molarity = $\frac{\text{no. of moles}}{volume}$

                                          = $\frac{2.17 \times 10^{-7} mol}{1 L}$

                                          = $2.17 \times 10^{-7}$ mol/L

Since, for the given solution Molarity = osmolarity

Thus, we can conclude that osmolarity of .00001 grams (0.1 mg%) of ethanol  in 1 L is $2.17 \times 10^{-7}$ osmol/L.