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(H2O)=24mm of Hg ) is x×10−2mm of Hg, the value of x is .......... .
=9.16×10−3mol ‌‌=0.00916mol Number of moles of NaOH in the solution, n2=‌
100×0.1
100
=1×10−2mol=0.01mol
Here, NaOH is the limiting reagent as it will leave (0.00916−0.005) mole of SO2 after the reaction. So, number of moles of Na2SO3 (solute) produced in the solution, n2′=‌
1
2
×0.01=0.005mol It is added into 36g of water to observe the colligative property, Relative Lowering of Vapour Pressure (RLVP). No. of moles of solute (Na2SO3)=0.005mol=n2′ No. of moles of solvent (H2O)=36∕18=2mol=n1‌′ For Na2SO3 van't Hoff factor, i=3, as Na2SO3⟶2Na++SO32−‌‌[α=1, strong electrolyte ] i=[1+α(n−1)]=[1+1×(3−1)]=3 The RLVP equation is ‌
∆p
p∘
=x‌solute ‌′×i=‌
n2′
n1′+n2′
×i Lowering of VP (of the solution) =‌
n2′
n1′+n2′
×i×p∘=‌
0.005
2+0.005
×3×24mm of Hg =0.1795∼eq0.18mm of Hg=18×10−2mm of Hg =x×10−2mm of Hg⇒x=18