Question

A 11.3 mL sample of CH3OH (d = 0.793 g/mL) is dissolved in enough water to produce 75.0 mL of a solution with a density of 0.980 g/mL. What is the solution

concentration expressed as ; (a) molarity of CH3OH; (b) molality of CH3OH?

Answer 1

A) molarity= moles of solute/Liters of solution

m(CH3OH)= d×V= 0.793×11.3=8.96g

n(CH3OH)= m/M= 8.96/32.04= 0.280 mol

Molarity(M)= 0.280 mol/0.0750 L= 3.73 M

b) molality(m)=moles of solute/kilograms if solvent

m(solution)= m(solute)+m(solvent)

m(solution)=d×V= 75.0×0.980= 73.5g

m(solute)= m(CH3OH)= 8.96 g

m(solvent)= 73.5-8.96= 64.5 g

molality(m)= 0.280 mol/0.0645 kg= 4.34 m