Question

What is the concentration (m) of ch3oh in a solution prepared by dissolving 12.9 g of ch3oh in sufficient water to give exactly 230 ml of solution?

Answer 1

Concentration (m) signify 'molality'

Concentration (M) signify 'Molarity'

Formula for molality (m) =
`(Moles of solute)/(Mass of solvent in Kg)`

In the question denisty of solute is not given so we can calculate concentration (M) that is 'Molarity'

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

Moles of solute CH3OH = Given grams / Molar mass of CH3OH

Given grams of CH3OH = 12.9 g and molar mass = 32.0 g/mol

Moles of solute CH3OH =
`12.9 g* (1mol)/(32.0g)`

Moles of solute CH3OH = 0.403 mol

Volume of solution = 230 mL , we need to convert 230 mL to 'L'

Volume =
`230 mL* (1 L)/(1000 mL)`

Volume = 0.23 L

Molarity (M) =
`(0.403mol)/(0.23L)`

Concentration (M) = 1.75 mol / L or 1.75 M