Sol (1)
SO20.0821×298224=9.2mmol+2NaOH10mmol(L.R.)→Na2SO35mmol(i=3)+H2O
P0−Ps=iXsoluteP0
=24×23×5×10−3
=0.18=18×10−3 mm of Hg
224mL of SO2(g) at 298K and 1atm is passed through 100mL of 0.1MNaOH solution. The non-volatile solute produced is dissolved in 36g of water. The lowering of vapour pressure of solution (assuming the solution is dilute) (P∘(H2O)=24mm of Hg) is x×10−2mm of Hg the value of x is (Integer answer)
Held on 26 Feb 2021 · Verified 6 Jul 2026.
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