Title 2. SOLUTIONS - Explanations.pdf ✅

1 (b) (π) glucose =(π) unknown compound 0.05= 3 M M = 3 0.05 = 60 n= 60 30 =2 (e.f.m. for CH2O = 30) so, molecular formula=C2H4O2 2 (a) △ Tf = ikfm where △ Tf = depression in freezing point i=van,t Hoff factor m= molality and and kf= freezing point depression constant For 0.01 molal NaCl solution 0.37 = 2 × kf × 0.01 ∴ kf = 0.37 2×0.01 -----(i) For 0.02 molal urea solution △ Tf = 1 × kf × 0.02 ∴ kf = △Tf 0.02 -----(ii) From Eqs (i) and (ii) 0.37 2×0.01 = △Tf 0.02 △ Tf = 0.37×0.02 2×0.01 ∴ △ Tf = 0.37∘C 3 (b) Na2SO4 ⇌ 2Na + + SO4 2− van’t Hoff factor i=[1+(y-1) α] where y is the number of ions from one mole solute, (in this case =3), α the degree of dissociation. i=(1+2 α) 4 (c) N = w × 1000 eq. wt.× V(mL) = 10 × 1000 60 × 100 = 1.66 N 6 (a) Molality = 18 180 = 0.1 molal 7 (a) PN2 = KH × mole − fraction (N2) mole-fraction (N2 ) 1 105 × 0.8 × 5 = 4 × 10−5mol −1 In 10 mole solubility is 4 × 10−4 . 8 (d) PT=PA ∘XA + PB ∘xB Mixture solution boil at 1 atm = 760 mm = total pressure. 760 = 520 XA + 100(1 − XA) XA = 0.5, mol% of A =50% 9 (b) Depression in freezing point is colligative property. The solute which produces highest number of ions will have minimum freezing point . 1. One molal NaCl aqueous solution NaCl → Na + + Cl − ∴ 2 ions/molecule 2. One molal CaCl2 solution CaCl2 → Ca 2+ + 2Cl − ∴ 3 ions/molecule 3. One molal KCl aqueous solution KCl → K + + Cl − ∴ 2 ions/molecule 4. One molal urea aqueous solution →no dissociation ∵ CaCl2 solution has highest number of ions ∴ It has lowest freezing point. 10 (a) Ideal solution △H =0 △ V =0 FA−A = FB−B= FA−B 11 (c) Given, vapour pressure of benzene, p ∘=640 mm Hg Vapour pressure of solution, p=600 mm Hg Weight of solute, w= 2.175 g Weight of benzene, W= 39.08 g Molecular weight of benzene, M=78 g Molecular weight of solute, m=? According to Raoult’s law, P ∘−P P∘ = w×M m×W 640−600 640 = 2.175×78 m×39.08 40 640 = 2.175×78 m×39.08 m= 16×2.175×78 39.08 m=69.60 12 (b) N = w × 1000 eq. wt.× V(mL) = 4 × 1000 40 × 100 = 1.0 N 13 (a) Van’t Hoff’s factor (i)=4 {3K +[Fe(CN)6 ] 3−} Molality = 0.1 329 × 1000 100 = 1 329 ⟹ -△ Tf = iKf. m
= 4 × 1.86 × 1 329 = 2.3 × 10−2 ⟹ Tf = −2.3 × 10−2°C (As freezing point of water is 0 ∘C) 14 (b) ∆x = i × kf × m 7.10 × 10−3 = i × 1.86 × 0.001 i = 3.817 ∵ α = i − 1 n − 1 ∴ 1 = 3.817 − 1 (x + 1) − 1 x = 2.817 ≈ 3 ∴ molecular formula of the compound is K3[Fe(CN)6] 15 (b) For NaCl, i = 2 ∆Tf = 2kf × m = 2 × 1.86 × 1 = 3.72 Ts = T − ∆Tf = 0 − 3.72 = −3.72°C 16 (c) Number of moles = Molarity × Volume (in L) ⟹ Number of moles of H2 SO4 = 2.0 M × 5.0 L = 10 moles 17 (c) We know that 1 g equivalent weight of NaOH = 40 g ∴ 40 g of NaOH = 1 g eq. Of NaOH ∴ 0.275 g of NaOH = 1 40 ×0.275 eq. = 1 40 × 0.275 × 1000 =6.88 meq ∴ N1V1 = N2V2 (HCl) (NAOH) N1 × 35.4 = 6.88 (∵ meq = NV) N1 = 0.194 18 (d) Equivalent weight of K2Cr2O7 = molecular weighty of K2Cr2O7 oxidation number of Cr Oxidation number of Cr in K2Cr2O7 2[+1]+2(x)+7(-2)= 0 2+2x-14 = 0 2x=12 x=6 Equivalent weight =294.19 6 = 49.08 weight of K2Cr2O7 equivalent wt.(E) = N × V(L) w = 0.1 × 1 × 49.03 = 4.903 g 20 (a) According to Raoult’s law p = pA ∘XA + pB ∘ XB = 290 = 200 × 0.4 + pB ∘ × 0.6 pB ∘ = 350 21 (c) Two solutions are isotonic if their osmotic pressure are equal. π1 = π2 M1ST1 = M2ST2 (M1and M2 are molarities) At a given temperature, M1 = M2 1000w1 m1V1 = 1000w2 m2V2 (V1 = V2 = 100mL) Cane sugar unkown ∴ w1 m1 = w2 m2 5 329 = 1 m2 m2 = 342 5 =68.4 g mol −1 22 (d) According to Raoult’s law, PA = PA ∘ κA or κA = PA PA ∘ = 32mm Hg 40 mm Hg = 0.8 23 (a) One molar (1 M) aqueous solution is more concentrated than one molal aqueous solution of the same solute. In solution, H2SO4 provides three ions. While NaCl provides two ions. Hence, vapour pressure of solution of NaCl is higher (as it gives less ions). Therefore, 1 molal NaCl will have the maximum vapour pressure. 24 (b) The number of moles or gram molecules of solute dissolved in 1000 g of solvent = molality 117 g NaCl = 2 mol Hence, concentration of solution = 2 molal 25 (a) In solution the KCl and CuSO4 produces same number of ions in solution. KCl ⇌ K + + Cl− CuSO4 ⇌ Cu2+ + SO4 2− Both produced two ions in solution. So, ionic strength of a solution is combined ionic strength of both of the salt. =0.1+.02=0.3 mol/kg 26 (b) △ Tf = i × kf × m HBr ⟶ H + + Br − Ions at equilibrium 1-α α α ∴ Total ions = 1 − α + α + α = 1 + α ∴ i = 1 + α Given, kf = 1.86 K mol −1 Mass of HBr = 8.1 g Mass of H2O =100 g (α) = degree of ionization = 90% m(molality) =mass of solute mol.wt.of solute mass of solvent in kg
= 8.1/81 100/1000 i = 1 + α =1+90/100 =1.9 △ Tf = i × kf × m = 1.9× 1.86 × 8.1 /81 100/1000 = 3.534 ∘C △ Tf =(depression in freezing point) = freezing point of water - freezing point of solution 3.534 = 0 – freezing point of solution. ∴ Freezing point of solution = -3.534 ∘C 27 (d) M= 1000×kf×w ∆Tf×W = 1000×1.86×4.5 0.465×100 = 180 g 28 (c) p 0 − ps p 0 = w m × M w 0.30 mm 17.54 mm = 20 m × 18 100 ⇒ m = 20 × 18 × 17.54 0.30 × 100 = 210.48 29 (c) For ideal solution, △ Hsolution =△ H1 +△ H2 +△ H3 31 (c) Molarity = Moles of solute Volume of solution(L) moles of urea =120 60 = 2 weight of solution =weight of solution + weight of solute =1000 + 120 =1120 g ⇒ Volume = 1120g 1.15g mL × 1 1000mL/L =0.974 K ⇒Molarity = 2.000 0.974 = 2.05 M 32 (a) π =CRT Hence, C =0.2 M R =0.082 L atm mol −1K −1 T =27+273 = 300 K π =0.2× 0.082 × 300 K =4.92 atm. 33 (c) Molarity of base = Normality Acidity = 0.1 1 = 0.1 M1V1 = M2V2 0.1 × 19.85 = M2 × 20 M2 = 0.09925 ≈ 0.099 34 (a) Molarity = normality × equivalent weight molecular weight Given, normality of Na2CO2 solution =0.2 N Equivalent weight = M Molecular weight 2 M (∵ Na2CO3 is dipositive.) ∴ Molarity = 0.2 × M 2M = 0.1 M 35 (b) Given ps =19.8 mm nA = 0.1 nB = 178.2 18 = 9.9 According to Raoult’s law ps−p ps = nA nA+nB 19.8−p 19.8 = 0.1 9.9+0.1 or 198-10 p = 19.8 ×0.1 10 p = 198-1.98 10 p =196.02 p = 19.602 mm 36 (a) N1V1 = N2V2 36 × 50 = N2 × 100 ∴ N2 = 36 × 50 100 = 18 ∵ Molarity of acid = Normality Basicity = 18 2 = 9 M 37 (b) ∆Tf = i × kf × m i for HBr=1 +α where, α =degree of dissociation i=1+0.9=1.9 ∴ ∆Tf = 1.9 ×1.86× 8.1×1000 100×81 =3.534°C Freezing point =-3.534°C 38 (d) Moles = mass molecular weight Given, mass of Al2 (SO4)3 = 50 g molecular mass of Al2 (SO4)3 = 342 ∴ Moles of Al2 (SO4)3 = 50 342 = 0.14 mol 39 (b) Let the volume of 0.4 M HCl is V1 and that of 0.9 M HCl is V2. We know that, NV = N1V1 + N2 V2 (Mixture) (for 0.4 M HCl) ( for 0.9 M HCl) 0.7(V1 + V2) = 0.4 × V1 + 0.9 × V2 [∵1m HCl = 1N HCl] 0.7V1 + 0.7V2 = 0.4 V1 + 0.9 V2 0.7V1 + 0.4V1 = 0.9V2 + 0.7V2 0.3V1 = 0.2V2 V1 V2 = 0.2 0.3 = 2 3 40 (b)
n-heptane and ethanol forms non-ideal solution. In pure ethanol, Molecules are hydrogen bonded. On adding n-heptane, its molecules get in between the host molecules and break sme of the hydrogen bonds between them. Due to weaking of interactions, the solution shows positive deviation from Raoult’s law. 41 (a) W = NE V 1000 N = W×1000 E×V = 6.3×1000 63×250 =0.4N N1V1 = N2V2 0.1 × V1 = 0.4 × 10 V1 = 0.4 × 10 0.1 V1 = 40 mL 42 (c) Ca(NO3)2 ⇋ Ca 2+ + 2NO3 − It furnishes 3 ions per formula unit. So, its van’t Hoff factor is 3. 43 (b) π = CRT ∶ C = π RT = 7.8 0.0821 × 310 = 0.31 mol/L 44 (a) When ethylene glycol is added to H2O as antifreeze, it decreases the freezing point of H2O in winter and increase the boiling point of water in the summer. 45 (b) Molarity of H2SO4 =5 M Normality of H2SO4 = 2 × 5 =10 N N1V1 = N2V2 10 × 1 = N2 × 10 or N2 =1 N 46 (a) Molality, m= no.of moles of solute weight of solution in kg = 1000×w1 m1W1 = 1000×0.6 60×200 = 0.05 [∴ Molecular weight of NH2CONH2 = 60] Given, ∆Tb = 0.05 ∆Tb = Kb × m or 0.05 =Kb × 0.05 ∴ Kb = 10 K mol −1 47 (b) Molarity = Number of moles of solute Volume of solution (in L) ⟹ molarity = 5 205 =2M 48 (b) pH = −log[H +] log[H +] = −pH = 0.00 [H +]=antilog(0.00) [H +]= 1.0 M M H2SO4 = 2NH2SO4 ∴ Normality of 250mL solution =2×250 1000 =0.50 N 49 (a) AxBy ⇌ xA y+ + yB x− After dissociation (1-α) xα yα i=n(AxBy)+n(A y+) + n(B x− ) =2-α+x α + yα = 1 + α(x + y − 1) ∴ α = i−1 (x+y−1) 50 (a) w =0.15 g, W=15 g, ∆Tb =0.216°C kb = 2.16°C ∴ M=kb×w×100 ∆Tb×W = 2.16×0.15×1000 0.216×15 = 100 51 (d) Osmotic pressure of two solutions will be added. Hence, osmotic pressure of resulting solution =1.64+2.46 =4.10 atm. 52 (a) According to molarity equation NaOH = HCl M1V1 = M2V2 0.6 × V1 = 0.4 × 30 V1 = 0.4 × 30 0.6 = 20cm3 53 (c) Mole fraction and molality does not involve volume therefore they are independent of temperature. 54 (a) For complete neutralisation, m. wq of H2SO4 = m. eq. of NaOH 0.1 × 2 × V = 50 × 0.2 × 1 (∵ 0.1M H2SO4 = 0.2N H2SO4) V = 50mL 55 (b) ∆Tb = mkb ∆Tf = mkf ∆Tb ∆Tf = kb kf = 0.512 1.86 ∆Tb = 0.512 1.86 × 0.186 =0.0512° 56 (a) Molarity = of solution × 10(in litre) M where, M = molecular weight of the solute Molarity = 40 × 1.2 × 10 M ×1000 ...(i) Molarity = weight of the solute /M volume of solution (in litre) ...(ii) From Eqs. (i)and(ii)