The ionic equation given here is 3SO2 + Cr2O7 (2- ) + 2H (+) --> 3SO4 (2- ) + 2Cr (3+) + H20.

The polluted air contains an amount of SO2 that occupies 4.8% of 3.0 dm^3 of the air. For every three molecules of SO2 we require 1 ion...

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The ionic equation given here is 3SO2 + Cr2O7 (2- ) + 2H (+) --> 3SO4 (2- ) + 2Cr (3+) + H20.

The polluted air contains an amount of SO2 that occupies 4.8% of 3.0 dm^3 of the air. For every three molecules of SO2 we require 1 ion of Cr2O7 (2- ).

At STP, using PV = nRT, the moles of SO2 in the air is given by:

n = PV / RT

=> n = [4.8%*100*1000*(3/1000)]/ 273*8.3145

=> n = 4.8*3/273*8.3145

=> n = 6.34 * 10^-3 mol.

So we need the dichromate ions for 6.34 * 10^-3 moles of SO2.

We need 2.11* 10^-3 mol of dichromate ion.

Now there is 1 mole of dichromate ions in 10 dm^3 of solution or in 10/1000 m^3 of the solution.

2.11*10^-3 moles will be found in 2.11*10^-3*10^-2 m^3 of the solution.

**The minimum volume of dichromate solution required is 2.11*10^-5 m^3.**