JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Solutions

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 is 36 g of water. The lowering of vapour pressure of solution (assuming the solution is dilute)

(P_{(H_{2} O)}^{\circ} = 24 mm

of Hg

)

x \times 1 0^{- 2} mm

of Hg, the value of x is ______. (Integer answer)

Answer: 24

Solution

At the stated conditions the

224 mL

S O_{2}

\frac{224}{22400} = 0.01

mol, and the alkali supplies

0.1 \times 0.1 = 0.01

mol of

NaOH

. Because the gas and the base are present in equal amounts, the reaction stops at the acid salt,

S O_{2} + NaOH \to NaHS O_{3}

, giving

0.01

mol of sodium hydrogensulphite. This salt is a strong electrolyte and separates into

N a^{+}

and

HS O_{3}^{-}

, so the number of dissolved particles is

2 \times 0.01 = 0.02

mol, while the water amounts to

\frac{36}{18} = 2

mol. For a dilute solution the relative lowering of vapour pressure equals the mole fraction of the dissolved particles, so

Δ p = p^{\circ} \frac{0.02}{2} = 24 \times 0.01 = 0.24 mm Hg .

Written in the form

x \times 1 0^{- 2} mm Hg

this is

24 \times 1 0^{- 2}

, so

x = 24

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Integer answer

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JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Solutions

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