A commercially available sample of sulphuric acid is 15 percent H2SO4 by weight Density 1.10g/ml. Calculate molality of the solution??

Molarity = $\frac{number of moles of solute}{volume of solution (in mL)} \times 1000$

15% of H₂SO₄ means 15 gm in 85 gm of solvent i.e. water

Number of moles = mass/molar mass

mass of H₂SO₄ = 98 g/mol

Therefore, Number of moles of H₂SO₄ = 15/98

                                                                = 0.15 moles

Density of solution = Mass/Volume

Mass of solution = 100 gm

Density of solution = 1.10 g/mL

Volume = Mass/Density

             = 100/1.10

             = 90.9 mL
So,
     Molarity(M) = $\frac{0.15 \times 1000}{90.9}$

                          = 1500/909

                          = 1.68 Molar (or just M)
Now,
        Molality(m) = $\frac{number of moles of solute}{mass of solvent (in gm)} \times 1000$

                            = $\frac{0.15\times 1000}{85}$

                            = 150/85

        Molality(m) = 1.7 molal (or just m)

Hope this information clears your doubts about the topic.

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