1. 0.5g fuming H₂SO₄ is diluted with water.the solution requires 26.7ml of 0.4N NaOH for complete neutralisation.find the % of free SO₃ in the sample?

Ans:20.8%

2. a mixture of 2 immiscible liquids nitrobenzene and water boiling at 99⁰C has a partial pressure of water 733mm and that of nitrobenzene 27mm.calculate ratio of weights of nitrobenzene to water in the distillate.

Ans:1/4

1) fuming H2SO4 = H2SO4 + SO3

let mass of H2SO4 = x, SO3 = 0.5-x

when diluted with water SO3 turns into H2SO4.

So, extra H2SO4 = (0.5-x)/80 * 98   ( no of moles of SO3 = 0.5-x / 80)

After dilution, total H2SO4 = (49 - 18x)/80 g.

apply N1V1 = N2V2.

no of millieq of H2SO4 = 26.7 * 0.4 = 10.68

so amount of H2SO4 = 10.68 * 49 *10^-3 = 0.52332.

Now, (49-18x)/80 = 0.52332.

x=0.396

Free SO3 in the sample = 0.5-x = 0.104

% = 20.8

2) mole fractions of water = 733/760 = 0.964

                                    Nitrobenzene = 0.036

mass of nitrobenzene = 0.036 * 123 = 4.428

mass of water = 0.96 4* 18 = 17.36

mass ratio = 0.25 (approx)

here x=0.104 then how come its 20.8% and not 10.4% according to your solution.

plz explain me this step

So, extra H2SO4 = (0.5-x)/80 * 98   ( no of moles of SO3 = 0.5-x / 80)

and this

After dilution, total H2SO4 = (49 - 18x)/80 g.

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