Question

what would be the molality of a solution made by mixing equal volume of 30% by mass of H2SO4 ( density = 1.218 g/cc) and 70% by mass of H2SO4 ( density = 1.61 g/cc) ?

Answer 1

H2SO4= 98 

    weight of 30% of H2SO4 soln = 1.218 g                               = 1000 ml * 1.218 g/ml =1.218g (v*d)

    wt. of H2SO4 (solute) = 1218*30/100 = 365.4 g

    wt. of H20 (solvent) = 1218-365.4 

    now , total wt. of H2SO4 = 365.4+1127 = 1492.4 g     
   70% of H2SO4 SOln = 1000 ml * 1.61 g/cc =1610g

    wt. of H2SO4 (solute) = 1610*70/100 = 1127 g

    wt. of H2O (solvent) = 1610-1127 = 483 g
    now, tota wt. of H2O (Solvent)= 852.6 + 483 = 1335.6 g 
  finalllllyyyy, molarity = wt*100/molecular wt.* wt. of solvent                           = 1492.4*1000/98*1335.6 = 11.4 

Answer 2

Answer:

Molarity : 7.60M & Molality : 11.38m

here's the answer and hope it helps you.