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Q.

224 mL of SO2g at 298 K and 1 atm is passed through 100 mL of 0.1 M NaOH solution. The non-volatile solute produced is dissolved in 36 g of water. The lowering of vapour Pressure of solution (assuming the solution is dilute) (P°H2O=24 mm of Hg) is x×102 mm of Hg, the value of x is ________ 

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Detailed Solution

 no. of moles of SO2n=PVRT =1×224×10-30.0821×298=9×10-3=9 milli moles  mill moles NaOH=0.1×100=10SO2+2NaOH(aq)Na2SO3+H2O

9                   10                        0

9-5             10-10                     5

No. of moles of  H2O=3618=2

Po-PsP0=3×5×10-32            Po-Ps =15×24×10-32 Po -Ps=18×10-2 

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