Question

Calculate molarity of a 20% w/w aqueous solution of sulphuric acid?

Answer 1

`"2.04 mol/L"`

Explanation:

`%"w/w" = "Mass of solute"/"Mass of solution" × 100`

`20%"w/w"` means `"20 g"` of sulfuric acid is present in `"100 g"` of water

Density of water = `"1 g/mL"`

Volume of water in given solution = `"100 g"/"1 g/mL" = "100 mL" = color(blue)"0.1 L"`

Molar mass of `"H"_2"SO"_4 = "98 g/mol"`

Moles of `"H"_2"SO"_4` in given solution = `"20 g"/"98 g/mol" = color(blue)(10/49"mol")`

`"Molarity" = "Moles of solute"/"Volume of solution (in litres)"`

`"Molarity" = (10/49"mol")/"0.1 L" = color(red)"2.04 mol/L"`

Answer 2

Relationship between `%"w/w"` and Molarity is

`"Molarity" = ("10" × %"w/w" × "density of solvent")/"Molar mass of solute"`

`"Molarity" = (10 × 20 × 1)/98 "mol/L" = "2.04 mol/L"`