Molality question help?

Question

A 2.45 M sulfuric acid solution, H2SO4, has a density of 1.15 g/mL at 20 degrees C. What is the molality at this temperature? Does molarity or molality change if we raise the temperature to 30 degrees C? Why?

I keep getting the wrong answer.. please explain, thanks!

Answer 1

2.45 mol/L x 98.078 g/mol = 240.3 g/L of H2SO4 in the solution

Mass water = 1150 g - 240.3 = 909.7 g => 0.9097 Kg

m = 2.45 mol / 0.9097 = 2.69

If we rise the temperature density will change and so M and m

Answer 2

Molarity is n (solute) / V (solution)

molality is n (solute) / m (solvent)

So 1 Lt of the solution you have will contain 2,45 moles of H2SO4 and will have a mass of 1,15 Kg.

2,45 moles of H2SO4 weight: 2,45 x 98,086 = 0,2403 Kg

Since the question asks you to get molality you will need the mass of the solvent that will be m (solution) - m (solute) = 1,15 - 0,2403 = 0,909 Kg

So the molality will be:   n (solute) / m (solvent) = 2,45 / 0,909 = 2,695 m

Only molarity will change with temperature arising because you need the volume of the solution and volume changes depending by temperature. Molality instead needs the mass of the solvent that remains the same at doesn't matter which temperature.