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

Molarity = $\frac{numberofmolesofsolute}{volumeofsolution(inmL)}\times 1000$

15% of H_{2}SO_{4} means 15 gm in 85 gm of solvent i.e. water

Number of moles = mass/molar mass

mass of H_{2}SO_{4} = 98 g/mol

Therefore, Number of moles of H_{2}SO_{4} = 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) = $\frac{numberofmolesofsolute}{massofsolvent(ingm)}\times 1000$

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

= 150/85

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

Hope this information clears your doubts about the topic.

Keep asking!!

Regards

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