94420 + Interview Questions in Chemistry in Class 12 Page 1747 InterviewSolution

87301.	A solution containing ethyl alcohol and propyl alcohol has a vapour pressure of 290mm at 30^(@)C. Find the vapour pressure of pure ethyl elcohol if its mole fraction in the solution is 0.65. The vapour pressure of propyl alcohol is 210mm at the same temperature.
Answer» ANSWER :`333.12mm`

Discussion

87302.	A solution containing Cu(I), Ni and Zn cyanide complexes was electrolysed and a deposit of 0.175 g was obtained. The deposit contained 72.8% Cu, 4.3% Ni and 22.9% Zn. No other element was released. Calculate the number of coulombs passed through the solution.
Answer» ANSWER :335.9 C

Discussion

87303.	A solution containing compound X in water and a solution containing urea in water were put in a closed system. By doing this some water vapour was removed from one solution and got condensed in the other. It is found that when both the solutions were at equilibrium vapour pressure, one solution contains 2% of X and the other 5% by weight. Find the molecular weight of X.
Answer» SOLUTION : At EQUILIBRIUM, the RELATIVE lowering of vapour pressure of the TWO solutionsis equal. 23.26

Discussion

87304.	A solution containing a group-IV cation gives a precipitate on passing, H_(2)S. A solution of this precipitate in dil. HCl produces a white precipitate with NaOH solution and bluish-white prcipitate with basic potassium ferrocyanide. The cation is :
Answer» `MN^(2+)` `ZN^(2+)` `NI^(2+)` `Co^(2+)` Answer :B

Discussion

87305.	A solution containing 8.6g urea in one litre was found be isotonic with a 5%(wt./vol.) solution of an organic non-volatile solute molecular weight of latter is :
Answer» ANSWER :A

Discussion

87306.	A solution containing 7.45 g of KCI per litre of solution has an osmotic pressure of 4.68 atm 300 K. Calculate the degree of dissociation of KCI in solution.
Answer» Solution :Calculation of observed MOLAR mass of KCI `"According to Van't Hoff equation ":M_(B)=(W_(B)RT)/(PIV)` `W_(B)=7.45g, pi=4.68 atm, V=1 L, T=300 K, R= 0.0821" L atm K"^(-1)mol^(-1)` `M_(B)=((7.45g)xx(0.0821"L atm K"^(-1)mol^(-1))xx(300K))/((4.68" atm")xx(1L))=39.2" g mol"^(-1)` Step II. Calculation of Van't Hoff factor `i=("NORMAL molar mass")/("Observed molar mass")=((74.5"g mol"^(-1)))/((39.2"g mol"^(-1)))=1.9` Step III. `"Calculation of degree of dissociation "(alpha)` `alpha=(i-1)/(n-1)=(1.9-1)/(2-1)=0.9=0.9xx100=90%`.

Discussion

87307.	A solution containing 8.6 g urea in one litre was found to be isotonic with 0.5% (mass/vol) solution of an organic, non volatile solute. The molecular mass of organic non volatile solute is:
Answer» 348.9 g/mole 34/89 g/mole 3489 g/mole 3.489 g/mole ANSWER :B

Discussion

87308.	A solution containing 62.5 g of non-volatile solute per 1000 g of water hass a greezing point 1.06^(@)C less than tha of water. Calculate the molecular mass of the solute. (K_(f)=1.86 K kg mol^(-1))
Answer» SOLUTION :`W_(B)=62.5g, W_(A)=1 KG, DeltaT_(f) =1.06^(@)C=1.06 K` `K_(f)=1.86" K kg mol"^(-1), M_(B)=?` `M_(B)=(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))=((1.86" K kg mol"^(-1))xx(62.5 g))/((1.06 K)xx1Kg)=109.67" g mol"^(-1)`

Discussion

87309.	A solution containing 6 g of benzoic acid in 50 g ether (C_(2)H_(5)OC_(2)H_(5)) has a vapour pressure of 410 mm of mercury at 293 K. Given that the vapour pressure of ether at the same temperature is 442 mm of mercury, calculate the molecular mass of benzoic acid. (Assume that the solution is dilute).
Answer» <P> Solution :`(p^(@)-p_(s))/(p^(@))=(w_(2)//M_(2))/(w_(1)//M_(1))=(w_(2)M_(1))/(w_(1)M_(2))" ,"(442-410)/(442)=(6xx74)/(50xxM_(2))"or"M_(2)=(6xx74)/(50)XX(442)/(32)=122.65`

Discussion

87310.	A solution containing 6 g of benzoic acid, C_(6)H_(5)COOH in 50 g of diethy 1 ether, C_(2)H_(5)OC_(2)H_(5), has a vapour pressure of 410 mm Hg at 293 K. Calculate the molar mass of the benzoic acid. Vapour pressure of diethy1 eth at 293 K is 442 mm Hg.
Answer» Solution :`P_(A)^(@)=442" mm Hg",P_(S)=410" mm Hg", W_(B)=6 g , W_(A)=50G,` `M_(A)(C_(2)H_(5)-O-C_(2)H_(5))=74" g mol"^(-1)` According to Raoult's Law(solution is non-ideal), `(P_(A)^(@)-P_(S))/P_(S)=n_(B)/n_(A)=W_(B)/M_(B)xxM_(A)/W_(A)` `((442-410)mm)/(410 mm)=((6g))/((M_(B)))XX((74" g mol"^(-1)))/((50 g))` `M_(B)=((6g)xx(74" g mol"^(-1)))/((50g))xx((410 mm))/((32mm))=113.8" g mol"^(-1)`.

Discussion

87311.	A solution containing 6 g of a solute dissolved in 250 cm^(3) of water gave an osmotic pressure of 4.5 atmosphere at 27^(@)C. Calculate the boiling point of the solution. The molal elevation constant for water is 0.52^(@)C per 1000 g.
Answer» Solution :The data on osmotic pressure is : `w_(2)=6 g "" V=250 CM^(3)=0.25` litre `pi =4.5" atm," T = 27+273=300 K` R = 0.0821 litre atm/degree/mole van't Hoff equation for osmotic for osmotic pressure is : `pi V = nRT` `therefore 4.5xx0.25 = n xx0.0821xx300` or `n=(4.5xx0.25)/(0.0821xx300)=0.0457` mole i.e., 0.457 mole of the solute are present in 250 mL of water (or 250 g of water). `therefore` Molality of the solution `=(0.0457 mol)/(250 g)xx 1000 g kg^(-1)=0.1828" mol kg"^(-1)`, i.e., m = 0.1828 `K_(B)=0.52^(@)C//1000 "g (GIVEN)" therefore Delta T_(b)=K_(b)xx m=0.52xx0.1828=0.095^(@)C` `therefore` Boiling POINT of solution `(T_(b))=T_(b)^(@)+Delta T_(b)=100^(@)C+0.095^(@)=100.095^(@)C`

Discussion

87312.	A solution containing 50 g of ethaylene glycolin 200 g water is cooled to -9.3^(@)C. The amount of ice that will separate out will be (K_(f)="1.86 Km"^(-1))
Answer» 18.71 g 28.71 g 38.71 g 48.71 g Solution :We have to calculate mass of liquid WATER that is PRESENT in the solution. The remaining will freeze as ice. `DeltaT_(f)=(1000K_(f)w_(2))/(w_(1)M_(2))` `9.3=(1000xx1.86xx50)/(w_(1)xx62)[M_(2)" for "{:(CH_(2)OH),(\|),(CH_(2)OH):}=62]` `or w_(1)=161.29g` `THEREFORE"Ice separated out "=200-161.29=38.71g`

Discussion

87313.	A solution containing 500 g of a protein per litre is isotonic with a solution containing 3.42 g of sucrose per litre. The molecular mass of protein is:
Answer» 5 146 34200 50000 Answer :D

Discussion

87314.	A solution containing 4.5 mmol of Cr_2O_7^(2-) and 15 mmol of Cr^(3+) shows a pH of 2.0. calculate the potential of the half reaction Cr_2O_7^(2-) to Cr^(3+),E^@=1.33V
Answer» Solution :The complete half cell REACTION is `Cr_2O_7^(2-) +14H^+ +6e to 2Cr^(3+) +7H_2O, E^@=1.33V` Given: `[Cr_2O_7^(2-)]=4.5/1000M, [Cr^(3+)]=15/1000M and [H^+]=10^-2 M` `E=E^@-0.0591/6 log""[Cr^(3+)]^2/([Cr_2O_7^(2-)][H^+]^(14))` `E=1.33-0.0591/6 log""(0.015)^2/((0.0045)(10^-2)^(14))` `E=1.067V`

Discussion

87315.	A solution containing 4.2 g of urea in 500 ml was found to be isotonic with a 5% ( wt./vol.)solution of an organic non-volatile solute. The molar mass of the solute is
Answer» 35.71 325.8 357.1 418.6 Solution :For ISOTONIC SOLUTIONS `C_(1)=C_(2)` `(4.2xx1000)/(60xx500)=(5xx1000)/(Mxx100)` `:.M=(5xx500xx60)/(100xx4.2)` `=357.1`

Discussion

87316.	A solution containing 4.0 g of PVC in 2 litre of dioxane (industrial solvent ) was found to have an osmotic pressure 3.0xx10^(-4) atm at 27^(@)C The molecular mass of the polymer will be :
Answer» `1.6xx10^(4)` `1.6xx10^(5)` `1.6xx10^(3)` `1.6xx10^(2)` Answer :B

Discussion

87317.	A solution containing 4 g of polyvinyl chloride in 1 litre of dioxane was found to have an osmotic pressure of 6 xx 10^-4 atm at 300 K. The molecular mass of the polymer is:
Answer» `3 XX 10^3` `1.6 xx 10^5` `5 xx 10^4` `6.4 xx 10^2` ANSWER :B

Discussion

87318.	A solution containing 36 g of solute dissolved in one litre of water gave an osmotic pressure of 6.75 atmosphere at 27^(@)C. The molal elevation constant of water is 0.52^(@)C. Calculate the boiling point of the solution.
Answer» SOLUTION :`100.1425^(@)C`

Discussion

87319.	A solution containing 34.2 g of cane-sugar (C_(12)H_(22)O_(11)) dissolved in 500cm^(3) of water froze at -0.374^(@)C. Calculate the freezing point depression constant of water.
Answer» Solution :Here, we have`""DeltaT_(f)=0.374K` CONCENTRATION of sugar solution = 34.2 g in `500 cm^(3)="68.4 in 1000 cm"^(3)="68.4 in 1000 g"` `""(because" density of "H_(2)O="1 g cm"^(-3))` `=(68.4)/(342)" molal"(because "MOLAR mass of "C_(12)H_(22)O_(11)="342 g mol"^(-1))` `=0.2 m` `"APPLYING the relationship"DeltaT_(f)=K_(f)xxm, K_(f)=(DeltaT_(f))/(m)=(0.374 K)/(0.2m)="1.87 K m"^(-1)`

Discussion

87320.	A solution containing 4 g of a non-volatile organic solute per 100 ml was found to have an osmotic pressure equal to 500 cm of mercury at 27^@C. The molecular weight of solute is:
Answer» 14.97 149.7 1697 1.497 Answer :B

Discussion

87321.	A solution containing 3.3 g of a substance in 125g of benzene (b.p 80^(@)C) boils at 80.66^(@)C. If K_(b) for one litre of benzene is 3.28^(@)C, the molecular mass of the substance shall be
Answer» 127.2 131.2 137.12 142.72 Solution :`Delta T_(b) = (K_(b) XX W_(2))/(M_(2) xx W_(1))=1000` `:. M_(2) = (1000K_(b) W_(2))/(W_(1) Delta T_(b))=(1000 xx 3.28 xx 3.3)/(125 xx 0.66)=131.2`

Discussion

87322.	A solution containing 3.3g of a substance in 125g of benzene (b.pt. = 80^@C) boils at 80.66^@C. If K_b for benzene is 3.28 K kg mol l^(-1) the molecular mass of the substance will be:
Answer» `130.20` `129.20` `132.20` `131.20` ANSWER :D

Discussion

87323.	A solution containing 3.01 xx 10^(23)HCI molecules is diluted to a volume of 4 litres. The molar concentration of the solution is
Answer» 1 M 2 M 0.125 M 0.25M Solution :`3.01 xx 10^23 HCL`molecules=0.5 MOLE HCI MOLAR conc. `=(0.5"mol)/(4L)=0.125M`

Discussion

87324.	A solution containing 3.1 g of BaCI_(2)in this solution. (K_(b) for water= 0.52 K m^(-1). Molar mass BaCI_(2)=208.3 mol^(-1)).
Answer» Solution :`"Molality of solution (m)"=W_(B)/(M_(B)xxW_(A))` `W_(B)=3.1g, W_(A)=250 g =0.25 kg, M_(B)=208.3 " g MOL"^(-1)` `m=((3.1g))/((208.3" g mol"^(-1))xx(0.25 kg))=0.05952" kgmol"^(-1)=0.05952 m.` `"For the solution", Delata T_(b)=iK_(b)m` `or"" i=(DeltaT_(b))/(K_(b)xxm),DeltaT_(b)=100.083^(@)C-100^(@)C=0.083 K` `K_(b)=0.52Km^(-1),m=0.05952 m` `THEREFORE""i=((0.083 K))/((0.52"Km"^(-1))xx(0.05952m))=2.68` `"RATIO of i and m (i/m)"=(2.68)/(0.05952)=45:1`

Discussion

87325.	A solutioncontaining 3.100 g of BaCl_(2) in 250 gof waterboilsat 100.083^(@)C .Calculate the Van't Hoff factorand molality of BaCl_(2) in thissolution . (k_(b) for water = 0.52 Km^(-1) , molar mass of BaCl_(2) = 208.3 g mol^(-1))
Answer» Solution :MOLALITY of the solution , `m= (w_(B) xx 1000)/(M_(B) xx w_(A))` `w_(B) = 3.100 g, w_(A) = 250g, M_(B) = 208.3` `m = (3.100 xx 1000)/(208.3 xx 250) =0.05952` Now , LET US calculate normal elevationin boiling point , `Delta T_(b) = k_(b) xx m` ` = 0.05952 xx 0.52= 0.03095` Observed elevation in boiling point, `Delta T_(b) = 100.083 - 100 = 0.083^(@)C` ` i=("Observed" Delta T_(b))/("Normal"Delta T_(b)) = (0.083)/(0.03095) = 2.68`

Discussion

87326.	A solution containing 30 gms of non-volatile solute in exactly 90 gm water has a vapour pressure of 21.85 mm Hg at 25^(@)C. Further 18 gms of water is then added to the solution. The resulting solution has a vapour pressure of 22.15 mmHg at 25^(@)C. Calculatethe molecular weight of the solute
Answer» <P>74.2 75.6 67.83 78.7 Solution :We have, `(p^(@)-21.85)/(21.85)=(30xx18)/(90xx m)`, for 1 case…….(i) WT. of solvent `= 90+18=108 gm` `(p^(@)-22.15)/(22.15)=(30xx18)/(108xx m)`, for II case …..(ii) By EQ. (1) `p^(@)m-21.85m=21.85xx6=131.1` By eq. (2) `p^(@)m-22.15m =22.15xx5=110.75` `0.30m=20.35` `m=(20.35)/(0.30)=67.83`

Discussion

87327.	A solution containing 30 g of non - volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate :(i) Molar mass of the solute.(ii) Vapour pressure of water at 298 K.
Answer» Solution :(i) Let, the molar mass of the solute be M g `mol^(-1)` Now, the no. of moles of solvent (water), `n_(1)=(90g)/(18g mol^(-1))=5 mol` And, the no. of solute, `n_(2)=(30g)/(M mol^(-1))=(30)/(M)mol` `p_(1)=2.8` kPa Applying the relation: `(p_(1)^(0)p_(1))/(p_(1)^(0))=(n_(2))/(n_(1)+n_(2))` `therefore (p_(2)^(0)-2.8)/(p_(1)^(0)=((30)/(M))/(5+(30)/(M))` `therefore 1-(2.8)/(p_(1)^(0))=((30)/(M))/((5M+30)/(M))` `therefore 1-(2.8)/(p_(1)^(0))=(30)/(5M+30)` `therefore (2.8)/(p_(1)^(0))=1-(30)/(5M+30)` `therefore (28)/(p_(1)^(0))=(5M+30-30)/(5M+30)` `therefore (2.8)/(p_(1)^(0))=(5M)/(5M+30)` `therefore (p_(1)^(0))/(2.8)=(5M+30)/(5M) ""`....(i) After the addition of 18 g of water : `n_(1)=(90+18)/(18)=6mol` `p_(1)=2.9` kPa Aganin, applying the relation : `(p_(1)^(0)-p_(1))/(p_(1)^(0))=(n_(2))/(n_(1)+n_(2))` `therefore (p_(1)^(0)-2.9)/(p_(1)^(0))=((30)/(M))/(6+(30)/(M))` `therefore 1-(2.9)/(p_(1)^(0))=((30)/(M))/((6M+30)/(M))` `therefore 1-(2.9)/(p_(1)^(0))=(30)/(6M+30)` `therefore (2.9)/(p_(1)^(0))=1-(30)/(6M+30)` `therefore (2.9)/(p_(1)^(0))=(6M+30-30)/(6M+30)` `therefore (2.9)/(p_(1)^(0))=(6M)/(6M+30)` `therefore (p_(1)^(0))/(2.9)=(6M+30)/(6M) ""`....(ii) Dividing equation (i) or (ii), we have : `(2.9)/(2.8)=((5M+30)/(5M))/((6M+30)/(6M))` `(2.9)/(2.8)xx(6M+30)/(6)=(5M+30)/(5)` `2.9xx5xx(6M+30)=2.8xx6xx(5M+30)` `87M+435=84M+504` 3 M = 69 M = 23 Therefore, the molar mass of the solute is 23 g `mol^(-1)`. (ii) Putting the value of M in equation (i), we have : `(p_(1)^(0))/(2.8)=(5xx23+30)/(5xx23)` `therefore (p_(1)^(0))/(2.8)=(145)/(115)` `therefore p_(1)^(0)=3.53` kPa Hence, the VAPOUR pressure of water at 298 K is 3.53 kPa.

Discussion

87328.	A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further 18 g of water is then added to this solution, the new vapour pressure becomes 2.9 kPa at 298 K. Calculate (i) the molar mass of solute, (ii) vapour pressure of water at 298 K.
Answer» SOLUTION :Case I `(p_A^0 -p_A)/(p_A^0) = (w_B //M_B)/(w_A // M_A)` Substituting the VALUES, we have `(p_A^0 - 2.8)/(p_A^0) = (30 // M_B)/(90 //18) = (6)/(M_B) " or" (p_A^0 - 2.8)/(p_A^0) = (6)/(M_B)`....(i) Case II `(p_A^0 -2.9)/(p_A^0) = (30//M_B)/(108//18) = (5)/(M_B) " or " (p_A^0 - 2.9)/(p_A^0) = (5)/(M_B)`...(ii) DIVIDING equation (i) by (ii), we get `(p_A^0 - 2.8)/(p_A^0 - 2.9) = 6/5` On solving we get `p_A^0 = 3.4 kPa` ubstituting the VALUE of `p_A^0`in equation (i), we get `(3.4-2.8)/(3.4) = ( 6)/(M_B) " or " M_B = 34` g/mol

Discussion

87329.	A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressureof 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate : (i) molar mass of the solute(ii) vapour pressure of water at 298 K.
Answer» Solution : (i) First Experiment : Suppose the molar mass of the solute = Mg `"mol"^(-1 )` Number of MOLES, `n_2` (solute) = 30/M moles Number of moles, `n_1`(solvent, `H_2O`) =`(90g)/(10g "mol"^(-1) ) = `5 moles Using Raoult.s LAW EQUATION and substituting values, we have `(p^0 -p_s)/(p^0) = (n_2)/(n_1 + n_2) i.e., (p^0 -2.8)/(p^0) = (30//M)/(5 + 30//M)` ` 1- (2.8)/(p^0) = (30 //M)/(5 + 30 //M)` `(2.8)/(p^0) = 1 - (30 //M)/(5+ 30//M) = (5 + 30//M -30//M)/(5 + 30 //M) = (5)/(5+ 30 // M)` ` (p^0)(2.8) = (5 + 30//M)/(5) = 1 + (6)/(M)`..(1) Second Experiment : After adding 18 g of water. Number of moles of water `n_1` = 6 moles Substituting values in Raoult.s law equation, we have `(p^0- 2.9)/(p^0) = (30//M)/(6+ 30//M)` `1-(2.9)/(p^0) = (30 //M)/(6 + 30//M)` `(2.9)/(p^0) = 1 - (30 //M)/(6 + 30//M) = (6+ 30//M - 30 //M)/(6 + 30 //M) = (6)/(6+ 30//M)` `(p^0)/(2.9) = (6 + 30//M)/(6) = 1 + 5/M`....(2) There are TWO unknown quantities and two equations. Dividing equation (1) by equation (2), we get `(2.9)/(2.8) = (1+ 6//M)/(1 + 5//M) " or " 2.9 (1 + 5/M)= 2.8 (1+ 6/M)` `2.9 + (14.5)/(M) = 2.8+ (16.8)/(M) "or " (2.3)/(M) = 0.1 " or " M = 23 g "mol"^(-1)` (ii) Substituting M = 23 in equation (2), we get `(p^0)/(2.8) = 1 + 6/23 = 29/23` `p^0 = 29/23 xx 2.8 = 3.53 KPA`

Discussion

87330.	A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further 18 g of water is added to this solution. The new vapour pressure becomes 2.9 kPa at 298 K. Calculate (i) The molecular mass of solute and (ii) vapour pressure of water at 298K
Answer» Solution :Let the vapour pressure of water at 298 K = `P^(0)` the molecular Mass of solute = m the molecular Mass of water = 18 mass of water = 90 gm Mass of solute = 30 gm Mole fraction of the non-volatile solute = `(30//m)/((90)/(18) + (30)/(m)) = (30)/(30 + 5m) = 1 ` The vapour pressure of solute = 2.8 kPa at 298. Now, `"" = (P^(0) - 2.8)/(P^(0)) = (30)/(30 + 5m)`... (1) Now, 18 gm of water is ADDED to the solution Now , mass of water = 108 Mole fraction of the solute `= (30//m)/((180)/(18) + (30)/(m))= (30)/(30 + 6 m )` NEW relative lowering `(P^(0) - 2.9)/(p^(0))` `= (30)/(30 + 6m) "" ` ... (2) Soving equation (1) and (2) , we get m = 34 g Molecular mass of non-volatile solute = 34 g Vapour pressure of water at 298 K = 3.4 kPa

Discussion

87331.	A solution containing 30 g of a non-volatile solute exactly in 90 g water has a vapour pressure of 2.8 kPa at 298 K. Further 18 g of water is then added to the solution, the new vapour pressure becomes 2.9 kPa at 298 K. Calculate (i) molar mass of the solute. (ii) vapour pressure of water at 298 K.
Answer» Solution :Suppose the molar mass of the solute `="Mg mol"^(-1)` `n_(2)" (solute)"=(30)/(M)"moles,"n_(1) ("SOLVENT, "H_(2)O)=("90 g")/("18 g mol"^(-1))="5 moles"` `(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1)+n_(2)), i.e., (p^(@)-2.8)/(p^(@))=(30//M)/(5+30//M)or1-(2.8)/(p^(@))=(30//M)/(5+30//M)` `"or"(2.8)/(p^(@))=1-(30//M)/(5+30//M)=(5+30//M-30//M)/(5+30M)=(5)/(5+30//M)or (p^(@))/(2.8)=(5+30//M)/(5)=1+(6)/(M)"...(i)"` After adding 18 g of WATER, `n(H_(2)O)`, i.e., `n+(1)=6` moles `THEREFORE""(p^(@)-2.9)/(p^(@))=(30//M)/(6+30//M)or 1-(2.9)/(p^(@))=(30//M)/(6+30//M)` `"or"(2.9)/(p^(@))=1-(30//M)/(6+30//M)=(6+30//M-0//M)/(6+30//M)=(6)/(6+30//M)"or"(p^(@))/(2.9)=(6+30//M)/(6)=1+(5)/(M)"...(ii)"` Dividing eqn. (i) by eqn. (ii), we get `(2.9)/(2.8)=(1+6//M)/(1+5//M) "or"2.9(1+(5)/(M))=2.8(1+(6)/(M))` `"or"2.9+(14.5)/(M)=2.8+(16.8)/(M) "or"(2.3)/(M)=0.1 "or"M = 23u` (ii) PUTTING M = 23 in eqn. (i) we get `(p^(@))/(2.8)=1+(6)/(23)=(29)/(23)"or"p^(@)=(29)/(23)xx2.8="3.53 kPa."`

Discussion

87332.	A solution containing 2.675 g of CoCl_(3). 6 NH_(3) (molar mass = 267.5 g "mol"^(-1)) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of AgNO_(3) to give 4.78 g of AgCl (molar mass = 143.5 g "mol"^(-1)). The formula of the complex is (At. Mass of Ag = 108 u)
Answer» `[Co(NH_(3))_(6)]Cl_(3)` `[CoCl_(2)(NH_(3))_(4)]CL` `[COCL(NH_(3))_(3)]Cl_(3)` `[CoCl(NH_(3))_(5)]Cl_(2)` ANSWER :A

Discussion

87333.	A solution containing 2.675 g of CoCl_(3)*6NH_(3) (molar mass = 267.5 g mol^(-1)) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of AgNO_(3) to give 4.31 g of AgCl (molar mass = 143.5 g mol^(-1)). The formula of the complex is (Ag = 108 u)
Answer» `[Co(NH_(3))_(6)]Cl_(3)` `[CoCl_(2)(NH_(3))_(4)]Cl` `[CoCl_(3)(NH_(3))_(3)]` `[COCL(NH_(3))_(5)]Cl_(2)` SOLUTION :Mol of `Cl^(-)` = mol of AGCL = `4.31/143.5=0.03` `because` 0.01 mol of `CoCl_(3)6NH_(3)` produces 0.03 mol of `Cl^(-)` `therefore` 1 mole of `CoCl_(3)6NH_(3)` CONTAINS 3 `Cl^(-)` anions

Discussion

87334.	A solution containing 2.675 g of CoCl_(3).6NH_(3) (molar mass = 267.5 g "mol"^(-1)) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of AgNO_(3) to give 4.78 g of AgCl (molar mass = 143.5 g "mol"^(-1)). The formula of the complexis : (At. mass of Ag = 108u)
Answer» `[CoCl_(3)(NH_(3))_(3)]` `[CoCl(NH_(3))_(5)]Cl_(2)` `[Co(NH_(3))_(6)]Cl_(3)` `[CoCl_(2)(NH_(3))_(4)]Cl`. Solution :Moles of complex `=(2.675)/(267.5)=0.01` Moles of AgCl precipitated `=(4.78)/(143.5)=0.03` `therefore Cl^(-)` IONS produced = 0.03 moles. It shows that 3 moles of `Cl^(-)` ions are produced per mole of complex. `therefore` Possible structure of complex is `[Co(NH_(3))_(6)]Cl_(3)`. `UNDERSET(0.01"mol")underset(1"mol")([Co(NH_(3))_(6)]Cl_(3)) to [Co(NH_(3))_(6)]^(3+) +underset(0.03"mol")underset(3"mol")(3Cl^(-))`

Discussion

87335.	A solution containing 2.665 g of CrСl_(3).6H_(2)O is passed through a cation exchanger.The chloride ions obtained in solution react with AgNO_(3) and give 2.87 g of AgCI. Determine the structure of the compound.
Answer» Solution :`143.45 g` of AgCl contains 35.45 g Cl" ions therefore 2.879 of A` CI_(-)` will contain `(35.45xx 2.87)/(143.45)=o.709 GCL^(-)ions` `CrCl_(3).6H_(2)0` contains `n xx 35.45` g of ionizable CR ions (where n = no. of.`Cl^(-)` ions outside the coordination SPHERE). thus, 2.665 g `ClCl_(3)6H_(2)O` will contain `(NXX 35.45 xx2.665)/(266.35)` Also `nxx 35.45xx 2.665)/(266.35)= 0.709 implies napprox2` Keeping in view the octahedral geometry of the complex, its structure may be written as `[CrCI(_(2)O)_(5),Cl_(2)H_(2)O]`.

Discussion

87336.	A solution containing 2.665 g of CrCl_3.6H_(2)O is passed through a cation exchanger. The chloride ions obtained in solution react with AgNO_3 and 2.87g of AgCl_3 is precipitated. Determine the structure of the complex. ( Cr = 52 , Cl = 35.5 , Ag = 108 , N = 14 )
Answer» Solution :Amount of `CrCl_3.6H_(2)O` REACTED = 2.665 g Molar mass of the complex = `(52 + 3 xx 35.5 + 6 xx 18)` = `266.5 g mol^(-1)` No of moles of the complex reacted = `(2.665 g)/ (266.5 g mol^(-1)` = 0.01 mol Amount of AGCL formed = 2.87 g Molar mass of AgCl = `(108 + 35.5)` = `143.5 g mol^(-1)` Noof moles of AgCl formed = `(2.87 g ) / (143.5 g mol^(-1)` = 0.02 mol 0.01 mole of the complex gives AgCl = 0.02 mol 1 mole of the complex gives AgCl = 2 mol The number of free `CL^(-)` ions = 2 C.N of Cr in the complex = 6 `therefore` the complex may be represented as : `[CrClH_(2)O_(5)]Cl_(2).H_(2)O` : pentaaquachloridochromium (III) chloride monohydrate.

Discussion

87337.	A solution containing 25.6 g of sulphur, dissolved in 1000 g of naphthalene whose melting point is 80.1^(@)C gave a freezing point lowering of 0.680^(@)C. Calculate the formula of sulphur (K_(f) for napthalene =6.8 K m^(-1))
Answer» Solution :`W_(B)=25.6g, W_(A)=1 kg, DeltaT_(f)=0.68K, K_(f)=6.8" k kg mol"^(-1)` `M_(B)=(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))=((6.8" K kg mol"^(-1))xx(25.6 g))/((0.68K)xx(1kg))=256" g mol"^(-1)` `"Atomicity of sulphur"=((256" g mol"^(-1)))/((32" g mol"^(-1)))=8` `"THEREFORE, molecular fomula of sulphur"=S_(8)`.

Discussion

87338.	A solution containing 25.6gm of sulphur dissolved in 1000gm of naphthalene gave a freezing point lowering of 0.680, then molecular formula of sulphur is (K_F for naphthalene 6.8K kg "mol"^(-1))
Answer» `S_2` `S_4` `S_6` `S_8` ANSWER :D

Discussion

87339.	A solution containing 2.56 g of sulphur dissolved in 100 g of naphthalene whose melting point is 80^(@)1C gave a freezing point lowering of 0.680^(@)C. Calculate the formula of sulphur (K_(f) for naphthalene = 6.8 K/m)
Answer» Solution :`w_(2)="2.56 G,"w_(1)="100 g ,"K_(f)="6.8 Km"^(-1), DeltaT_(f)=0.68^(@)C` `"Now,"M_(2)=(1000K_(f)w_(2))/(w_(1)xxDeltaT_(f))` Substituting the value of `K_(f), w_(2), w_(1) and DeltaT_(f), M_(2)=("1000 g KG"^(-1)xx"6.8 K kg mol"^(-1)xx"2.56 g")/("100 g "xx"0.68 K")="256 g mol"^(-1)` LET molecular formula of sulphur be `S_(x)" ,"` Atomic mass of sulphur = 32 Molecular mass of `S_(x)=x xx 32""therefore""x xx32=256"or"x=(256)/(32)=8` Therefore, the molecular formula of sulphur is `S_(8).`

Discussion

87340.	A solution containing 25. 6 gm of sulphur dissolved in 1000 gm of napthalene gave a freezing point lowering of 0.680, then molecular formula of sulphare is (K_(f) for naphthalene=6.8K kg mol^(-1))
Answer» `s_(2)` `s_(4)` `S_(6)` `S_(8)` ANSWER :D

Discussion

87341.	A solution containing 2.08 g of anhydrours barium chloride is 400 CC of water has a specific conductivity 0.0058 ohm^(–1) cm^(–1). What are molar and equivalent conductivities of this solution.
Answer» ANSWER :A::B::C

Discussion

87342.	A solution containing 20.0 grams of a nonvolatile solute in exactly 1.00 mole of volatile solvent has a vapoure pressure of 0.500 atm at 20^(@)C. A second mole of solvet is added to the mixture, and the resulting solution has a vapoure pressure of 0.550 at of 20^(@)C. What is the molecular weight of the solute ?( g/mole)
Answer» `45.0` `90.0` `20.0` `50.0` ANSWER :B

Discussion

87343.	If electrode potential of following cell: Pt_(s)\|Fe_(aq)^(2+),Fe_(aq)^(3+)\|\|MnO_(4(aq))^(-),Mn_(aq)^(2+),H_(aq)^(+)\|Pt_(s) is X then calculate value of 20X. [Given: E_(MnO_(4)^(-)\|Mn^(2+) = 1.51 V, E_(Fe^(3+)\|Fe^(2+) = 0.78 V,(2.303 RT)/F = 0.06]
Answer» SOLUTION :`-0.12 - (0.0591)/2 log (1/X) = -0.24` `log 1/X = (0.12 XX 2)/0.06 =4` `X = 10^(-4)`

Discussion

87344.	A solution containing 18g of non - volatile non - electrolyte solute is dissolved in 200g of water freezes at 272.07K. Calculate the molecular mass of solute. Given K_(f)=1.86kg//mol and freezing point of water = 273K
Answer» Solution :`Delta T_(f) = K_(f)xx m or (K_(f) xx W_(2) xx 1000)/(M_(2) xx W_(1))` `M_(2) = (1.86 xx 18 xx 1000)/((273- 272.07) xx 200)` = 180 g/mol.

Discussion

87345.	A solution containing 1.8 g of a compound (empirical formula CH_(2)O) in 40 g of water is observed to freeze at -0.465^(@)C. The molecular formula of the compound is (K_(f) of water = 1.86 kg k "mol"^(-1))
Answer» `C_(2)H_(4)O_(2)` `C_(3)H_(6)` `C_(4)H_(8)O_(4)` `C_(6)H_(12)O_(6)` Solution :`M_(2)=(1000K_(f)w_(2))/(w_(1)xxDeltaT_(f))=(1000xx1.86xx1.8)/(40xx0.465)=180` `"E.F. mass "CH_(2)O=30 THEREFORE n=(180)/(30)=6` `therefore"Molecular formula "=6xxCH_(2)O=C_(6)H_(12)O_(6)`

Discussion

87346.	A solution containing 18 g of anon-volitile solute in 200 g of waer freezes at 272.07 K. Calculate the molecular mass of the solute. (Given K_(f)=1.86 K/m)
Answer» Solution :`W_(B)=18g, W_(A)=0.2Kg, K_(f)=1.86," K kg MOL"^(-1), T_(f)=273 K, T_(f)=272.07 K` `DeltaT_(f)=T_(f)^(@)-T_(f)=273-272.07=0.93 K, M_(B)=?` `M_(B)=(K_(f)xxW_(B))/(DeltaT_(f)xxW_(A))=((1.86" K kg mol"^(-1))XX(18g))/((0.93K)xx(0.2 Kg))=180" g mol"^(-1).`

Discussion

87347.	A solution containing 1.8 g of a compound (empirical formula CH_(2)O) in 40 g os water is observed to freeze at -0.465^(@)C. The molecular formula of the compound is (K_(f) "of water" =1.86 kg K mol^(-1))
Answer» `C_(2)H_(4)O_(2)` `C_(3)H_(6)O_(3)` `C_(4)H_(8)O_(4)` `C_(6)H_(12)O_(6)` SOLUTION :`Delta T_(f)=K_(f)xxm` `0.465=1.86xx(1.8)/(M)xx(1000)/(40)rArr M =180` MOLECULAR formula = `("EMPIRICAL formula")_(n)` `n=("Molecular mass")/("Empirical formula mass")=(180)/(30)=6` Molecular formaula `= (CH_(2)O)_(6)=C_(6)H_(12)O_(6)`.

Discussion

87348.	A solution containing 15 g of urea (Molar mass =60 g mol^(-1)) per litrea has the same osmotic presure (isotonic) as a solution of glucose (molar mass =180 g mol^(-1)) in water, Calculate themass of glucose present in one litre of the solution.
Answer» Solution :`"Osmotic pressure"(pi)=(W_(B)xxRxxT)/(M_(B)xxV)` `"For urea solution "(pi_(1))=((15g)xxRxxT)/((60" g mol"^(-1))XX(1L))` `"For glucose solution"(pi_(2))=(W_(B)RxxT)/((180" g mol"^(-1))xx(1L))` `"For isotinic solutions", pi_(1)=pi_(2)` `therefore((15g))/((60" g mol"^(-1)))=W_(B)/((180" g mol"^(-1)))or BW_(B)=((15g)xx(180" g mol"^(-1)))/((6.0" g mol"^(-1)))=45 g` `therefore "Mass of glucose"(W_(B))=45.0 g.`

Discussion

87349.	A solution containing 12.5 g of non-electrolyte substance in 175 g of water gave boiling point elevation of 0.70 K. Calculate the molar mass of th substance (K_(b) for water=0.52 K kg mol^(-1)).
Answer» Solution :`M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xx W_(A))` `W_(B)=12.5g , W_(A)=175.0g=0.175 kg, DeltaT_(b)=0.70 K` `K_(b)=0.52" k kg MOL"^(-1)` `M_(B)=((12.5 g)xx(0.52" kg mol"^(-1)))/((0.70" K")xx(0.175" kg"))=53.06" g mol"^(-1)`.

Discussion

87350.	A solution containing 1.23g of Ca(NO_(3))_(2) in 10g of water boils at 100.975^(@)C. Calculate the degree of ionisation of the nitrate (K_(b)=0.52)
Answer» ANSWER :`75%`

Discussion

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A solution containing 25.6gm of sulphur dissolved in 1000gm of naphthalene gave a freezing point lowering of 0.680, then molecular formula of sulphur is (K_F for naphthalene 6.8K kg "mol"^(-1))

A solution containing 2.56 g of sulphur dissolved in 100 g of naphthalene whose melting point is 80^(@)1C gave a freezing point lowering of 0.680^(@)C. Calculate the formula of sulphur (K_(f) for naphthalene = 6.8 K/m)

A solution containing 25. 6 gm of sulphur dissolved in 1000 gm of napthalene gave a freezing point lowering of 0.680, then molecular formula of sulphare is (K_(f) for naphthalene=6.8K kg mol^(-1))

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A solution containing 18g of non - volatile non - electrolyte solute is dissolved in 200g of water freezes at 272.07K. Calculate the molecular mass of solute. Given K_(f)=1.86kg//mol and freezing point of water = 273K

A solution containing 1.8 g of a compound (empirical formula CH_(2)O) in 40 g of water is observed to freeze at -0.465^(@)C. The molecular formula of the compound is (K_(f) of water = 1.86 kg k "mol"^(-1))

A solution containing 18 g of anon-volitile solute in 200 g of waer freezes at 272.07 K. Calculate the molecular mass of the solute. (Given K_(f)=1.86 K/m)

A solution containing 1.8 g of a compound (empirical formula CH_(2)O) in 40 g os water is observed to freeze at -0.465^(@)C. The molecular formula of the compound is (K_(f) "of water" =1.86 kg K mol^(-1))

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A solution containing 12.5 g of non-electrolyte substance in 175 g of water gave boiling point elevation of 0.70 K. Calculate the molar mass of th substance (K_(b) for water=0.52 K kg mol^(-1)).

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