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#