In the aqueous solution of sulfuric …
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Gaurav Seth 6 years, 8 months ago
Mole fraction of solution = solute + solvent = 1
Here, Mole fraction of water is 0.85 so mole fraction of sulphuric acid will be (1-0.85) = 0.15
Hence each mole of the solution, 0.15 mole of sulphuric acid is dissolved .
So mass of solvent (water) is ,
For 1 mole of water H2O = 18 gm
So, 0.85 mole of water = 18 x 0.85 = 15.3 gm =0.0153 Kg
Thus, Molality (m) = no. of moles of solute (Sulphuric acid)/ mass of solvent in Kg
Hence, Molality of the solution (m) = 0.15 / 0.0153 = 9.8 m (nearly equal to 10)
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