224mL of SO_2(g) at 298K and 1 atm passes through 100mL of 0.1M NaOH solution. The non-volatile solute produced is dissolved is 36g of water. The lowering of vapour pressure of solution (assuming the solution is dilute) $(P_{(H_{2} O)}^{o} = 24 m m o f H g) i s x * 1 0^{- 2} m m$ of Hg, the value of x is-----------. (Integer answer)

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Payal Gupta | Contributor-Level 10

7 months ago

Moles of SO₂= $\frac{224 * 1 0^{- 3}}{22.4}$

=0.01 mole

Moles of NaOH = 0.1 * 0.1

= 0.01 mole

SO₂+NaOHNaHSO₃

0.01 mole0.01 mole-

-0.01 mole

Non-volatile solute is NaHSO3

Moles of water = $\frac{36}{18} = 2$

Using ; relative lowering in V.P

$\frac{P^{o} - P}{P^{o}} = i x_{B}$

Where; $Δ P = P^{0} - P$ is lowering in V.P

$Δ P = P^{0} i x_{B}$

i for NaHSO₃ = 2

here; $x_{B} = \frac{n_{B}}{n_{A}}$ since solution is dilute

$Δ P = 24 * 2 * \frac{0.01}{2} = 0.24$

$Δ P = 24 * 1 0^{- 2} m m H g$

So; x = 24

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