You are supplied with 25% (by mass) aqueous solution of Sulphuric acid

1.	You are supplied with 25% (by mass) aqueous solution of Sulphuric acid(98g/mol) having a density of 1.84g/cc. (3a)Calculate the molality of the solution. (show calculation)
Answer» Given:- → Concentration [by mass] = 25% → Molar mass of Sulphuric acid = 98g/mol → Density of the solution = 1.84g/cc To find:- → Molality of the solution. Solution:- It is clear that sulphuric acid is solute and water is solvent. Let the TOTAL mass of the solution be 100g. Then mass of SOLUTE [sulphuric acid] :- = 100/100 × 25 = 25g And mass of solvent [water] :- = 100 - 25 = 75g Number of mole in 25g of solute :- = Given Mass/Molar mass = 20/98 = 0.255 mole [approximately] Now, let's convert the mass of solvent from 'grams' to 'kilograms' :- => 1G = 0.001kg => 75g = 75(0.001) => 0.075kg Molality of a solution :- = Moles of solute/Mass of solvent in kg = 0.255/0.075 = 255/75 = 3.4m Thus, molality of the solution is 3.4m .

You are supplied with 25% (by mass) aqueous solution of Sulphuric acid(98g/mol) having a density of 1.84g/cc. (3a)Calculate the molality of the solution. (show calculation)

Answer»

→ Concentration [by mass] = 25%

→ Molar mass of Sulphuric acid = 98g/mol

→ Density of the solution = 1.84g/cc

→ Molality of the solution.

It is clear that sulphuric acid is solute and water is solvent.

Let the TOTAL mass of the solution be 100g.

Then mass of SOLUTE [sulphuric acid] :-

= 100/100 × 25

= 25g

And mass of solvent [water] :-

= 100 - 25

= 75g

Number of mole in 25g of solute :-

= Given Mass/Molar mass

= 20/98

= 0.255 mole [approximately]

Now, let's convert the mass of solvent from 'grams' to 'kilograms' :-

=> 1G = 0.001kg

=> 75g = 75(0.001)

=> 0.075kg

Molality of a solution :-

= Moles of solute/Mass of solvent in kg

= 0.255/0.075

= 255/75

= 3.4m

Thus, molality of the solution is 3.4m .