Solutions

(1) If molar mass of C₆H₅COOH is Mg mol^$-$1, then no. of moles of dissolved C₆H₅COOH = ${{1.22} \over M}$ mol

$\therefore$ Molality (m) of the solution = ${{1.22} \over M} \times {{1000} \over {100}} = {{12.2} \over M}$ mol kg^$-$1

We know, $\Delta$T_b = k_b $\times$ m

Given : $\Delta$T_b = 0.17 K and k_b = 1.7 K kg mol^$-$1

$\therefore$ $0.17 = 1.7 \times {{12.2} \over M}$ or, M = 122

$\therefore$ Molecular mass of C₆H₅COOH is = 122

(2) If molar mass of C₆H₅COOH is M', then no. of moles of dissolved C₆H₅COOH = ${{1.22} \over {M'}}$ mol

Molality (m) of the solution = ${{1.22} \over {M'}} \times {{1000} \over {100}} = {{12.2} \over {M'}}$ mol kg^$-$1

We know, $\Delta$T_b = 0.13 K and k_b = 2.6 K kg mol^$-$1

$\therefore$ $0.13 = 2.6 \times {{12.2} \over {M'}}$ or, M' = 244

$\therefore$ Molecular weight of C₆H₅COOH = 122

Now, van't Hoff factor (i) = ${{Calculated\,molecular\,weight} \over {Experimental\,molecular\,weight}} = {{122} \over {244}} = {1 \over 2}$

$i = {1 \over 2}$ indicates that C₆H₅COOH undergoes complete dimerisation in benzene.

Mass of water = 500 $\times$ 0.997 = 498.5 g and that of acetic acid = 3 $\times$ 10^$-$3 kg = 3 g.

No. of moles of acetic acid in solution = ${3 \over {60}} = 0.05$ mol [$\because$ Molar mass of acetic acid = 60 g mol^$-$1]

Molality (m) of the solution = ${{0.05} \over {498.5}} \times 1000 = 0.1$ mol kg^$-$1

Given : $\alpha$ = 0.23. Now, 1 molecule of CH₃COOH on dissociation in aqueous solution produces 2 ions,

CH₃COOH(aq) $\to$ CH₃COO^$-$(aq) + H⁺(aq)

So, n = 2

$\therefore$ $0.23 = {{i - 1} \over {2 - 1}}$ or, i = 1.23

For a solution of an electrolyte, the freezing point depression of the solution is given by, $\Delta$T_f = i $\times$ k_f $\times$ m

$\therefore$ $\Delta$T_f = 1.23 $\times$ 1.86 $\times$ 0.1 = 0.228 K

Hence, freezing point depression of solution = 0.228 K

As given in the question,

Determination of empirical formula of A :

$\therefore$ Empirical formula of minor product (A) = C₃H₂NO₂

Empirical formula mass = (12 $\times$ 3) + (1 $\times$ 2) + 14 $\times$ (16 $\times$ 2) = 84

Now, $\Delta$T_B = k_b $\times$ molality

or, $1.84 = 2.53 \times {{5.5} \over M} \times {1 \over {45}} \times 1000$

or, molar mass (M) = ${{2.53 \times 5.5 \times 1000} \over {1.84 \times 45}}$

$\therefore$ M = 168

$\therefore$ ${{molar\,mass} \over {formula\,mass}} = {{168} \over {84}} = 2$

$\therefore$ Molecular formula of (A) = (C₃H₂NO₂)₂ = C₆H₄N₂O₄

Structural formula :

We know, $\Delta$T_f = k_f $\times$ molality of solution (m)

Given, k_f for water = 1.86 K kg mol^$-$1

$\therefore$ 0.30 = 1.86 $\times$ molality of solution (m)

$\therefore$ m = ${{0.30} \over {1.86}}$ or, m = 0.161 mol kg^$-$1

According to Raoult's law, ${{{p^0} - p} \over {{p^0}}} = {x_2} \approx {{{n_2}} \over {{n_1}}}$

Amount of water in solution n₁ $\times$ 18 g

Molality of the solution ${{{n_2}} \over {{n_1} \times 18}} \times 1000 = {{55.55 \times {n_2}} \over {{n_1}}}$ mol kg^$-$1

But, molality calculated from f. p. depression = 0.161 mol kg^$-$1

$\therefore$ ${{55.55 \times {n_2}} \over {{n_1}}} = 0.0161$ or, ${{{n_2}} \over {{n_1}}} = 1.09 \times {10^{ - 3}}$

Given : p⁰ = 23.51 mm Hg

$\therefore$ ${{23.51 - p} \over {23.51}} = {{{n_2}} \over {{n_1}}} = 1.09 \times {10^{ - 3}}$

$\therefore$ p = 23.48 mm Hg

Volume of 1 mol of benzene $ = {{78} \over {0.877}} = 88.94$ mL and that for 1 mol of toluene $ = {{92} \over {0.867}} = 106.11$ mL

At 20$^\circ$C, volume of 1 mol of benzene vapour $ = 88.94 \times 2750 = 244585$ mL = 244.58 L & that for 1 mol of toluene vapour $ = 106.11 \times 7720 = 819.16$ L

The vapour pressure for pure benzene, $p_B^0 = {{nRT} \over V} = {{1 \times 0.0821 \times 293} \over {244.58}}$ atm = 0.098 atm and that for toluene, $p_T^0 = {{nRT} \over V} = {{1 \times 0.0821 \times 293} \over {819.16}}$ atm = 0.029 atm

Again, total vapour pressure = VP of benzene $\times$ mole fraction of benzene + VP of toluene $\times$ mole fraction of toluene

or, $P = p_B^0 \times {x_B} + p_T^0 \times {x_T}$ ...... [1]

Since, ${x_B} + {x_T} = 1$ $\therefore$ ${x_T} = 1 - {x_B}$ [x_B, x_T are mole fractions of benzene & toluene, respectively]

Given, total vapour pressure = 46 torr $ = {{46} \over {760}}$ atm = 0.060 atm

From equation [1], we get, $0.060 = 0.098 \times {x_B} + 0.029(1 - {x_B})$

or, $0.060 = 0.098{x_B} - 0.029{x_B} + 0.029$

or, ${x_B} = 0.45$ (in liquid phase)

$\therefore$ ${x_T} = 1 - 0.45 = 0.55$ (in liquid phase)

$\therefore$ Mole fraction of benzene in vapour phase $ = {{p_B^0 \times {x_B}} \over P} = {{0.098 \times 0.45} \over {0.060}} = 0.735$

Given, mass of water (W₁) = 100 g.

Let, the vapour pressure of water = p⁰

As given, vapour pressure of urea solution (p) $ = ({p^0} - 0.25{p^0}) = 0.75{p^0}$

According to Raoults' law, ${{{p^0} - p} \over {{p^0}}} = {{{W_2}/{M_2}} \over {{W_1}/{M_1} + {W_2}/{M_2}}}$

or, ${{{p^0} - 0.75{p^0}} \over {{p^0}}} = {{{W_2}/60} \over {100/18 + {W_2}/60}}$ [molar mass of urea = 60 g mol^$-$1 & that of water = 18 g mol^$-$1]

or, ${1 \over 4} = {{{W_2}/60} \over {{{1000 + 3{W_2}} \over {180}}}}$ or, ${W_2} = 111.1$

$\therefore$ 111.1 g of urea needs to be dissolved in the solution.

No. of moles of dissolved urea $ = {{111.1} \over {60}} = 1.85$ mol

Molality (m) of the solution $ = {{1.85} \over {100}} \times 1000 = 18.5$ mol kg^$-$1

We know, $\alpha = {{i - 1} \over {n - 1}}$

where $\alpha$ = degree of dissociation of an electrolyte in solution, n = no. of ions produced by 1 molecule of the electrolyte on dissociation, i = van't Hoff factor.

In aqueous solution, $Ca{(N{O_3})_2}$ dissociates as:

$Ca{(N{O_3})_2}(aq) \to C{a^{2 + }}(aq) + 2NO_3^ - (aq)$

Therefore, n = 3 and given : $\alpha$ = 0.7

$\therefore$ $0.7 = {{i - 1} \over {3 - 1}}$ $\therefore$ $i = 2.4$

For an electrolyte solution, the relative lowering vapour pressure is given by,

${{{p^0} - p} \over {{p^0}}} = i \times {x_2} \approx i \times {{{n_2}} \over {{n_1}}} = 2.4 \times {{{n_2}} \over {{n_1}}}$

In solution, no. of moles of $Ca{(N{O_3})_2}({n_2}) = {7 \over {1.64}} = 0.042$ mol [$\therefore$ Molar mass of $Ca{(N{O_3})_2}$ = 164 g mol^$-$1] and that of water $({n_1}) = {{100} \over {18}} = 5.55$ mol

Given : p⁰ = 760 mm Hg

$\therefore$ ${{760 - p} \over {760}} = {{2.4 \times 0.042} \over {5.55}}$ $\therefore$ p = 746.2 mm Hg

$\therefore$ Vapour pressure of the solution = 746.2 mm Hg

Given, vapour pressure of pure benzene, p⁰ = 640 mm Hg, vapour pressure of the solution, p = 600 mm Hg, mass of benzene, W₁ = 39.0 g, mass of the solid, W₂ = 2.175 g

We know, M₁ = molar mass of benzene = 78.

According to Raoults' law, ${{{p^0} - p} \over {{p^0}}} = {{{W_2} \times {M_1}} \over {{M_2} \times {W_1}}}$

$\therefore$ ${{640 - 600} \over {640}} = {{2.175 \times 78} \over {{M_2} \times 39}}$ or, ${M_2} = {{2.175 \times 78 \times 640} \over {39 \times 40}}$

or, ${M_2} = 69.6$, i.e., molecular mass of the solid = 69.6.

Given, at 373 K vapour pressure of the aqueous solution of glucose, p = 750 mm Hg.

We know, boiling point of water = 373 K. At this temperature, vapour pressure of water, p0
= 760 mm Hg

According to Raoults' law, ${{{p^0} - p} \over {{p^0}}} = {x_2} \approx {{{n_2}} \over {{n_1}}}$

$\therefore$ ${{760 - 750} \over {760}} = {{{n_2}} \over {{n_1}}}$ or, ${n_2} = {n_1} \times 0.013$

(1) Amount of water in solution n₁ mol = 18 $\times$ n₁g

$\therefore$ Molality of the solution $ = {{{n_2} \times 1000} \over {18 \times {n_1}}} = {{{n_1} \times 0.013 \times 1000} \over {18 \times {n_1}}}$ = 0.72 mol kg^$-$1

(2) Mole fraction of solute in solution $ = {{{n_2}} \over {{n_1} + {n_2}}} = {{{n_1} \times 0.013} \over {{n_1} + {n_1} \times 0.013}} = 0.0128$

Total vapour pressure (P) of an ideal solution made up of ethanol and methanol is given by, $P = {x_1}p_1^0 + {x_2}p_2^0$ where $p_1^0$ & $p_2^0$ are the vapour pressures of pure ethanol and pure methanol respectively and x₁ and x₂ are the mole fractions of ethanol and methanol respectively in solution.

Now, number of moles of ethanol $ = {{60} \over {46}} = 1.3$ (Molar mass of ethanol = 46 g mol^$-$1

and that of methanol $ = {{40} \over {32}} = 1.25$ (Molar mass of methanol = 32 g mol^$-$1

Total moles of ethanol & methanol = (1.3 + 1.25) = 2.55 mol

$\therefore$ ${x_1} = {{1.3} \over {2.55}} = 0.50$ and ${x_2} = {{1.25} \over {2.55}} = 0.49$

Given : $p_1^0$ = 44.5 mm Hg and $p_2^0$ = 88.7 mm Hg

$\therefore$ $p = {x_1}p_1^0 + {x_2}p_2^0 = (0.50 \times 44.5 + 0.49 \times 88.7)$ mm Hg = 65.71 mm Hg

$\therefore$ Total vapour pressure of the solution = 65.71 mm Hg

mole fraction of methanol in vapour $ = {{Partial\,v.p.\,of\,methanol\,in\,vapour\,of\,the\,solution} \over {Total\,v.p.\,of\,solution}}$

$ = {{{x_2}p_2^0} \over P} = {{0.49 \times 88.7} \over {65.71}} = 0.66$

Given : ${C_x}{H_{2y}}{O_y} + 2x{O_2} \to xC{O_2} + y{H_2}O + x{O_2}$

After cooling, only CO₂ and O₂ will be present in gaseous state because water exists in liquid form at 0$^\circ$C and 1 atm.

$\therefore$ Number of moles of gases after cooling = x + x = 2x

Volume of gases after cooling = 2.24 L (given)

$\therefore$ Moles of gases after cooling $ = {{2.24} \over {22.4}} = 0.1$

$\therefore$ 0.1 = 2x, or, x = 0.05

As given in the question, water collected during cooling = 0.9 g

$\therefore$ Number of moles of water $ = {{0.9} \over {18}} = 0.05$; hence, y = 0.05

$\therefore$ The empirical formula of the organic compound is CH₂O.

Given, vapour pressure of pure water (p⁰) = 17.5 mm Hg

Lowering of vapour pressure (p⁰ $-$ p) = 0.104 mm Hg

No. of moles of water in solution $({n_1}) = {{1000} \over {18}} = 55.55$ mol of the molar mass of solute is Mg mol^$-$1, then the number of moles of solute in solution $({n_2}) = {{50} \over M}$ mol

According to Raoults' law, ${{{p^0} - p} \over {{p^0}}} = {x_2} \approx {{{n_2}} \over {{n_1}}}$

$\therefore$ ${{0.104} \over {17.5}} = {{50} \over {M \times 55.55}}$ or, $M = {{50 \times 17.5} \over {55.55 \times 0.104}}$ or, M = 151.4

$\therefore$ Molar mass of the organic compound = 151.4 g mol^$-$1

Now, empirical formula mass of CH₂O = 30 g mol^$-$1

$\therefore$ $n = {{Molecular\,mass} \over {Empirical\,formula\,mass}} = {{151.4} \over {30}}$ or, $n = 5.04 \simeq 5$

$\therefore$ Molecular formula $ = {[C({H_2}O)]_5} = {C_5}{H_{10}}{O_5}$

As given, vapour pressure of pure benzene, p⁰ = 639.7 mm Hg and that of solution, p = 631.9 mm Hg.

If, n₁ and n₂ be the number of moles of benzene and solute respectively, then according to Raoults' law,

${{{p^0} - p} \over {{p^0}}} = {x_2} \approx {{{n_2}} \over {{n_1}}}$ or ${{639.7 - 631.9} \over {639.7}} = {{{n_2}} \over {{n_1}}}$

$\therefore$ ${n_2} = 0.0122 \times {n_1}$

Amount of benzene in solution = n₁ mol = 78 $\times$ n₁g [$\because$ Molar mass of benzene = 78 g mol^$-$1]

$\therefore$ Molality of the solution $ = {{{n_2} \times 1000} \over {78 \times {n_1}}} = {{0.0122 \times {n_1} \times 1000} \over {78 \times {n_1}}}$ = 0.156 mol kg^$-$1