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 .

Question

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 . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

= 9.2 m mol P s = P 0 ⋅ X text solvent =24 × (2/(2+15 × 10-3)) =23.82 Δ P =0.18 torr =18 × 10-2 torr

Δ P = P 0 ⋅ X text solute =24 × ((1.6+18.4)/2020) =0.2376=23.76 × 10-2

Q. 224mL of SO2(g)​ at 298K and 1 atm 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(H2​O)​=24mm of Hg) is x×10−2mm of Hg the value of x is ____.

=9.2mmol Ps=P0⋅Xsolvent ​ =24×(2+15×10−3)2​ =23.82 ΔP=0.18 torr =18×10−2 torr ΔP=P0⋅Xsolute ​ =24×2020(1.6+18.4)​ =0.2376=23.76×10−2

$= 9.2 m m o l$

$P^{s} = P^{0} \cdot X_{solvent}$

$= 24 \times \frac{2}{( 2 + 15 \times 1 0 ^{- 3} )}$

$= 23.82$

$Δ P = 0.18$ torr

$= 18 \times 1 0^{- 2}$ torr

$Δ P = P^{0} \cdot X_{solute}$

$= 24 \times \frac{( 1.6 + 18.4 )}{2020}$

$= 0.2376 = 23.76 \times 1 0^{- 2}$