Question

The free-base form of cocaine has a solubility of 1.00 g in 6.70 mL ethanol (CH3CH2OH). Calculate the molarity of a saturated solution of the free-base form of cocaine in ethanol.

Answer 1

Given information : Free-base form of cocaine solubility 1.00 g in 6.70 mL ethanol.

We have to find molarity.

`Molarity (M) =(Moles of solute)/(Volume of solution in L)`

To find molarity we need to first find moles of solute and volume of solution.

In our question solute is Cocaine(C17H21NO4) and ethanol (CH3CH2OH) is solvent.

Volume of solution = Volume of solute + Volume of solvent.

In our question , density of cocaine is not given that means amount of cocaine is very less as compared to that of solvent , so to find the volume of solution we can only consider volume of solvent that means we can take volume of solution as 6.70 mL.

Moles of solute cocaine (C17H21NO4) =
`(Grams of solute)/(Molar mass of solute)`

Molar mass of coacine (C17H21NO4) = 303.358 g/mol

Gram of solute given = 1 gram

`Moles of solute = 1 gramC17H21NO4* (1 mol C17H21NO4)/(303.358 gram C17H21NO4)`

Moles of solute C17H21NO4 = 0.00330 mol C17H21NO4

Volume of solution should be in L , so we will convert 6.70 mL to L

`Volume of solution = 6.70 mL* (1 L)/(1000 mL)`

Volume of solution = 0.0067 L

Now we can find molarity by plugging the moles of solute and volume of solution value in the formula.

`Molarity (M) =(Moles of solute)/(Volume of solution in L)`

`Molarity = ((0.00330 mol))/((0.0067 L))`

Molarity = 0.49 mol/L or 0.49 M