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

Question

Ellie Andrews · Accepted Answer

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# Consequently, molality is #m = n/(mass.solvent) = (0.288 mol es)/(0.250 kg) = 1.15m#

Noah Aldrich · Answer

undefined To find the molality, we first need to calculate the number of moles of solute (H2SO4) and then divide by the mass of the solvent (water) in kilograms.

1. Calculate the number of moles of H2SO4:
   moles = Molarity * Volume (in liters)
   moles = 6.00 mol/L * (48.0 mL / 1000 mL/L) = 0.288 mol

2. Calculate the mass of the solvent (water) in kilograms:
   mass = volume * density
   Since the density of water is approximately 1 g/mL, and the volume is 0.250 L:
   mass = 0.250 kg

3. Calculate the molality:
   molality = moles of solute / mass of solvent (in kg)
   molality = 0.288 mol / 0.250 kg = 1.15 mol/kg