Molarlity: An acqueous solution contains `128g` of mehanol `(CH_(2) OH)` in `108g` of water. Calculate the molarity of the solution.
Strategy: Convert the grams of `CH_(3) OH` to moles of `CH_(3) OH` express the mass of `H_(2) O` in kilogram and apply the definition of molality.

Explanation

Step 1: Calculate the moles of `CH_(3) OH`
Number of moles of `CH_(3) OH = ("Mass of" CH_(3) OH)/("Molar mass of" CH_(3) OH)`
`= (128g)/(32.0 g mol^(-1))`
`= 4.00 mol CH_(3) OH`
Step 2: Convert the mass of `H_(2) O` into kilograms.
Kilogram of `H_(2) O = (108g)/(1000g kg^(-1)) = 0.108 kg`
Step 3: Apply the definition of molality.
Molatiy `= ("Molar of solute")/("Mass of solvent(kg)")`
`= (4.00 mol)/(0.108 kg) = 37.0m CH_(3) OH`
Alternatively, in single setup,
`(? mol CH_(3) OH)/(kg H_(2) O) = (128g CH_(3) OH)/(0.108 kg H_(2) O) xx (1 mol CH_(3) OH)/(32.0g CH_(3) OH)`
`= (37.0 mol CH_(3) OH)/(kg H_(2) O)`
`= 37.0m CH_(2) OH`