What is the molality when 48.0 mL of 6.00 M H2SO4 are diluted into 0.250 L?

1 Answer
Dec 22, 2014

The answer is 1.15m.

Since molality is defined as moles of solute divided by kg of solvent, we need to calculated the moles of H_2SO_4 and the mass of the solvent, which I presume is water.

We can find the number of H_2SO_4 moles by using its molarity

C = n/V -> n_(H_2SO_4) = C * V_(H_2SO_4) = 6.00(mol es)/L * 48.0 * 10^(-3) L = 0.288

Since water has a density of 1.00 (kg)/L, the mass of solvent is

m = rho * V_(water) = 1.00 (kg)/L * 0.250 L = 0.250 kg

Therefore, molality is

m = n/(mass.solvent) = (0.288 mol es)/(0.250 kg) = 1.15m