Respuesta :
Given information : Free-base form of cocaine solubility 1.00 g in 6.70 mL ethanol.
We have to find molarity.
[tex] Molarity (M) =\frac{Moles of solute}{Volume of solution in L} [/tex]
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) = [tex] \frac{Grams of solute}{Molar mass of solute} [/tex]
Molar mass of coacine (C17H21NO4) = 303.358 g/mol
Gram of solute given = 1 gram
[tex] Moles of solute = 1 gramC17H21NO4\times \frac{1 mol C17H21NO4}{303.358 gram C17H21NO4} [/tex]
Moles of solute C17H21NO4 = 0.00330 mol C17H21NO4
Volume of solution should be in L , so we will convert 6.70 mL to L
[tex] Volume of solution = 6.70 mL\times \frac{1 L}{1000 mL} [/tex]
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.
[tex] Molarity (M) =\frac{Moles of solute}{Volume of solution in L} [/tex]
[tex] Molarity = \frac{(0.00330 mol)}{(0.0067 L)} [/tex]
Molarity = 0.49 mol/L or 0.49 M
