224   mL of SO 2 ( g ) at 298   K and 1   atm is passed through 100   mL of 0 . 1  …

224 mL of SO2(g) 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×10-2 mm of Hg the value of x is                (Integer answer)

Solution

Sol 1

SO2+2NaOH→  Na2SO3+H2O2240.0821×298=9.2 m mol10 mmol5 mmol(L.R.)(i=3)

P0-Ps=iXsolute P0

=24×3 ×5×10-32

=0.18 = 18×10-3 mm of Hg

Asked in: JEE Main 2021 (26 Feb Shift 1)

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