Question

Calculate molality for the following solutions. (Assume solution density is 0.998 g/mL) a) 0.540g Mg(NO3)2 in 250 mL of solution, b) 35 mL of 9.00 M H2SO4 diluted to 0.500 L

Answer 1

a)
Mass of Mg(No3)2  = 0.54 g
Molecular weight of the Mg(No3)2 = 148.3 g
Hence number of moles of Mg(No3)2 = 0.54/ 148.3 = 0.00364 moles

Molality = moles of solute ( in moles) / Mass of Solvent (in Kg)

Mass of solution = Volume * Density
   = 250 mL * 0.998 g/mL
= 249.5 g = 0.2495 kg
Mass of Solvent = Mass of Solution - Mass of Solute
   = 249.5 - 0.54 = 248.96
Molality = 0.00364/ 0.24896 = 0.0146 m


b)

When one solution is Diluted its Concentration changes.. and that can be given by

     
   35mL * 9 = 0.5L * M2
   ==> M2 = 0.63M
  
Molarity of the solution = moles of Solution / liter solution
it indicates that 0.63 moles of solution are there in 1L solution

The Mass of the solution = Volume * Density
   = 1L * 0.998 g/mL = 998 g
   Mass of the Solute = moles of solute * Molecular weight of the solute
   = 0.63 * 98 = 61.74 g
   Mass of the solvent = 998 - 61.74 = 936.26 g
   Molality = moles of solute/ Mass of solvent (in Kg)
   = 0.63/ (0.93626) = 0.672 m