A solution is prepared by dissolving 17.75 g sulfuric acid, h2so4, in enough water to make 100.0 ml of solution. if the density of the solution is 1.1094 g/ml, what is the molality?

Question

Corbin Reid

Chemistry

5 November, 15:08

A solution is prepared by dissolving 17.75 g sulfuric acid, h2so4, in enough water to make 100.0 ml of solution. if the density of the solution is 1.1094 g/ml, what is the molality?

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Cadence Frederick · Answer 1

When density = 1.1094 g / ml = 1.1094 Kg / L &

the volume of solvent = 100m/1000=0.1 L

weight of solvent = denisty x volume = 1.1094 kg/L * 0.1 L = 0.111 Kg

∴ molality = no. of moles of solute / Kg of solvent

we need to get the no. of moles of the solute = weight of solute/molar mass

= 17.75 / 98 = 0.18

by substitution in molality equation:

∴ molality = 0.18 / 0.111 = 1.62 mol/kg