Question

What would be the molality of solution made by mixing equal volumes of 30.0% by mass of \[{{H}_{2}}S{{O}_{4}}\] (density 1.218\[gc{{m}^{-3}}\]) and 70% by mass of \[{{H}_{2}}S{{O}_{4}}\] (density 1.1610\[gc{{m}^{-3}}\])?

Answer 1

Hint: For finding the molality we have to find the mass of solute, mass of solvent and moles of solute etc. because we know that molality is moles of solute per unit mass of solvent.

Step by step solution:
Molecular weight of \[{{H}_{2}}S{{O}_{4}}\]= 98
Let the volume of 1st solution is V\[c{{m}^{3}}\]
And given density of 1st solution \[{{d}_{1}}\]= 1.218\[gc{{m}^{-3}}\]
Weight of 30% of \[{{H}_{2}}S{{O}_{4}}\] solution = 1.218\[gc{{m}^{-3}}\]x V\[c{{m}^{3}}\]
                               = 1.218V g
w/w % = 30% (for 1st solution) = it means 30g of \[{{H}_{2}}S{{O}_{4}}\]in 100g of solution
100 g of solution = 30 g of\[{{H}_{2}}S{{O}_{4}}\]
1g of solution = (30/100) \[{{H}_{2}}S{{O}_{4}}\]
1.218V g of solution = \[\dfrac{30}{100}\times 1.218V\]\[{{H}_{2}}S{{O}_{4}}\]
                  = 0.3654Vg

Let the volume of the 2nd solution is also V\[c{{m}^{3}}\].
And given density of 2nd solution \[{{d}_{2}}\]= 1.1610\[gc{{m}^{-3}}\]
 Weight of 70% \[{{H}_{2}}S{{O}_{4}}\] solution = 1.1610\[gc{{m}^{-3}}\] x V\[c{{m}^{3}}\]= 1.1610V g
(w/w)% = 70% = it means 70 g of \[{{H}_{2}}S{{O}_{4}}\]in 100 g of solution
100 g of solution = 70 g of\[{{H}_{2}}S{{O}_{4}}\]
1g of solution = (70/100)g \[{{H}_{2}}S{{O}_{4}}\]
1.1610V g of solution = \[\dfrac{70}{100}\times 1.1610V\]g \[{{H}_{2}}S{{O}_{4}}\]
                    = 1.127V g
Now total mass of \[{{H}_{2}}S{{O}_{4}}\]= (0.3654Vg + 1.127V g)
                         = 1.4924V g
Moles of \[{{H}_{2}}S{{O}_{4}}\]= weight / molecular weight
                    = 1.4924V / 98
                    = 0.0152V mole
Now by mixing the both solutions the volume of total solution = V + V = 2V
Now molarity of new solution = moles / total volume
                           = 0.0152V / 2V

            = 0.0076 M

So, for molality
Total mass of solution = 1.218V + 1.1610V
                     = 2.828V g
Total mass of \[{{H}_{2}}S{{O}_{4}}\]= (0.3654Vg + 1.127V g)
                                    = 1.4924V g
Total mass of solvent = 2.828V - 1.4924V
                    = 1.3356V g = 0.0013356 kg
And we know that Molality = moles of solute / mass of solvent in (kg)
                     = \[\dfrac{0.0152V\text{ }mole}{0.0013356\text{ }kg}\]
               Molality = 11.38 m

Note: When you are doing calculation you should take volume as 2V not only V. And after mixing both solutions total moles will remain the same but the concentration of the final solution will change.