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) = $\frac{Moles of solute}{Mass of solvent in Kg}$

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

Molarity (M) = $\frac{Moles of solute}{Volume of solution in L}$

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 = $12.9 g\times \frac{1mol}{32.0g}$

Moles of solute CH3OH = 0.403 mol

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

Volume = $230 mL\times \frac{1 L}{1000 mL}$

Volume = 0.23 L

Molarity (M) = $\frac{0.403mol}{0.23L}$

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