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Mnenie [13.5K]
3 years ago
15

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

Chemistry
1 answer:
MrMuchimi3 years ago
8 0

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.

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Answer:

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It is clear that 2 mole of Na react with 1 mole of Cl₂ to  produce 2 moles of NaCl.

  • Firstly, we need to calculate the no. of moles of Na and Cl₂:

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  • From the stichiometry, Na reacts with Cl₂ with a molar ratio (2:1).

<em>So, 0.826 mol of Na "the limiting reactant" reacts completely with 0.413 mol of Cl₂ "left over reactant".</em>

The no. of moles of Cl₂ remained after the reaction = 0.48 mol - 0.413 mol = 0.067 mol.

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