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224 mL of SO 2 ( 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 ° H 2 O = 24 mm of Hg ) is x × 10 - 2 mm of Hg the value of x is (Integer answer)

Correct answer

0

Step-by-step solution

Sol 1 SO 2 + 2 NaOH →     Na 2 SO 3 + H 2 O 224 0 . 0821 × 298 = 9 . 2   m   mol 10   mmol 5   mmol ( L . R . ) ( i = 3 ) P 0 - P s = iX solute   P 0 = 24 × 3   × 5 × 10 - 3 2 = 0 . 18   =   18 × 10 - 3 mm of Hg

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