JEE MAIN - Chemistry (2021 - 26th February Morning Shift - No. 21)
224 mL of SO2(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 36g of water. The lowering of vapour pressure of solution (assuming the solution in dilute) (P$$_{({H_2}O)}^o$$ $$-$$ 24 mm of Hg) is x $$\times$$ 10$$-$$2 mm of Hg, the value of x is ___________. (Integer answer)
Answer
18TO24
Explanation
moles of SO2 = $${{224} \over {22400}}$$ = 0.01
moles of NaOH = molarity × volume (in litre)
= 0.1 × 0.1
= 0.01 moles
The balanced equation is
SO2 + 2NaOH $$ \to $$ Na2SO3 + H2O
$$ \therefore $$ Here NaOH is limiting Reagent.
2 mole NaOH produces 1 mole Na2SO3
0.01 mole NaOH produces $${1 \over 2}$$ $$ \times $$ 0.01 mole Na2SO3
$$ \therefore $$ Moles of Na2SO3 = 0.005 mole
Na2SO3 $$ \to $$ 2Na+ + SO32-
van’t Hoff factor (i) = 3
Moles of H2O = $${{36} \over {18}}$$ = 2 moles
Accoding to Relative Lowering of Vapour :
$${{P_{{H_2}O}^o - {P_S}} \over {P_{{H_2}O}^o}} = {{i{n_{N{a_2}C{O_3}}}} \over {{n_{{H_2}O}} + i{n_{N{a_2}C{O_3}}}}}$$
[ $${{n_{N{a_2}C{O_3}}}}$$ << $${{n_{{H_2}O}}}$$ ]
$${{P_{{H_2}O}^o - {P_S}} \over {P_{{H_2}O}^o}} = {{i{n_{N{a_2}C{O_3}}}} \over {{n_{{H_2}O}}}}$$
$$ \Rightarrow $$ $${{24 - {P_S}} \over {24}} = {{3 \times 0.005} \over 2}$$
$$ \Rightarrow $$ 24 – PS = 0.18
$$ \Rightarrow $$ PS = 23.82
Lowering in pressure ($$\Delta $$P) = 0.18 mm of Hg
= 18 × 10–2 mm of Hg
moles of NaOH = molarity × volume (in litre)
= 0.1 × 0.1
= 0.01 moles
The balanced equation is
SO2 + 2NaOH $$ \to $$ Na2SO3 + H2O
$$ \therefore $$ Here NaOH is limiting Reagent.
2 mole NaOH produces 1 mole Na2SO3
0.01 mole NaOH produces $${1 \over 2}$$ $$ \times $$ 0.01 mole Na2SO3
$$ \therefore $$ Moles of Na2SO3 = 0.005 mole
Na2SO3 $$ \to $$ 2Na+ + SO32-
van’t Hoff factor (i) = 3
Moles of H2O = $${{36} \over {18}}$$ = 2 moles
Accoding to Relative Lowering of Vapour :
$${{P_{{H_2}O}^o - {P_S}} \over {P_{{H_2}O}^o}} = {{i{n_{N{a_2}C{O_3}}}} \over {{n_{{H_2}O}} + i{n_{N{a_2}C{O_3}}}}}$$
[ $${{n_{N{a_2}C{O_3}}}}$$ << $${{n_{{H_2}O}}}$$ ]
$${{P_{{H_2}O}^o - {P_S}} \over {P_{{H_2}O}^o}} = {{i{n_{N{a_2}C{O_3}}}} \over {{n_{{H_2}O}}}}$$
$$ \Rightarrow $$ $${{24 - {P_S}} \over {24}} = {{3 \times 0.005} \over 2}$$
$$ \Rightarrow $$ 24 – PS = 0.18
$$ \Rightarrow $$ PS = 23.82
Lowering in pressure ($$\Delta $$P) = 0.18 mm of Hg
= 18 × 10–2 mm of Hg
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