94420 + Interview Questions in Chemistry in Class 12 Page 1856 InterviewSolution

92751.	1st I.P of nitrogen is higher than oxygen. Explain.
Answer» Answer :Due to presence of half filled orbital in NITROGEN which impart it extra STABILITY.

Discussion

92752.	"1m of MgCl_(2)" and 2 m of ura , aqueous solution are found to boilat same temperature . Hence we can say ________
Answer» `MgCl_(2)` is `50% ` ionised `MgCl_(2)` is `66.6%` ionised Both solutions have same vapour pressure Both solutionsfreeze at same temperature ANSWER :A::C::D

Discussion

92753.	1L of a buffer solution contains 0.1(M) CH_3COOH and 0.1(M) CH_3COONa. If 1 mL 10(M) HCl solution is added to this solution then find the change in the pH value. [pKa (CH_3COOH) = 4.74, change in volume can be neglected]
Answer» Solution :Before the ADDITION of HCl or NAOH `pH=14-(pK_b+log([NH_4Cl])/([NH_3]))` `=14-pK_b+log(0.1)/(0.1)=14-pK_b` After the addition of 0.02 mol of HCl `[NH_3]=0.1-0.02=0.08molL^(-1)` `[NH_3]=0.1+0.02=0.12 molL^(-1)` `:.pH=14-pK_b-log([NH_4Cl])/([NH_3])` `=14-pK_b-log(0.12)/(0.08)=13.824-pK_b` `:.` Decrease in pH`=14-pK_b-(13.824-pK_b)=0.176`

Discussion

92754.	1g pure iron is dissolved in excess of H_(2)SO_(4). The clear filtrate is made up 100 mL. 10mL of this solution is treated with 0.1 M KMnO_(4) solution till whole of the Fe^(2+) ions are oxidised to Fe^(3+) ions. Now 0.2 g Fe_(2)(SO_(4))_(3) is dissolved in it. the solution is now treated with Zn and H_(2)SO_(4). The ratio of equivalent of KMnO_(4) and K_(2)Cr_(2)O_(7) used for reducing the solution is:
Answer» `5//6` `6//5` `1` `2` SOLUTION :EQ. of `KMnO_(4)`=Eq. of `K_(2)Cr_(2)O_(7)`=Eq. of `FE^(2+)`

Discussion

92755.	1g pure iron is dissolved in excess of H_(2)SO_(4). The clear filtrate is made up 100 mL. 10mL of this solution is treated with 0.1 M KMnO_(4) solution till whole of the Fe^(2+) ions are oxidised to Fe^(3+) ions. Now 0.2 g Fe_(2)(SO_(4))_(3) is dissolved in it. the solution is now treated with Zn and H_(2)SO_(4). Select the correct statement. 1. The Fe^(3+) ions present in solution are reduced by Zn and H_(2)SO_(4) 2. H_(2) gas formed by the action of Zn and H_(2)SO_(4) is reducing agent. 3. Atomic form of H formed by the action of Zn and H_(2)SO_(4) is reducing agent 4. Nascent form of H formed by the action of Zn and H_(2)SO_(4) is reducing agent
Answer» `1,2` `1,3` `1,4` `1,2,3` Solution :`Zn+H_(2)SO_(4)toZnSO_(4)+underset("Nascent H")(2H)` `FE^(3+)+HtoFe^(2+)+H^(+)`

Discussion

92756.	1g pure iron is dissolved in excess of H_(2)SO_(4). The clear filtrate is made up 100 mL. 10mL of this solution is treated with 0.1 M KMnO_(4) solution till whole of the Fe^(2+) ions are oxidised to Fe^(3+) ions. Now 0.2 g Fe_(2)(SO_(4))_(3) is dissolved in it. the solution is now treated with Zn and H_(2)SO_(4). The volume of 0.1 M K_(2)Cr_(2)O_(7) used after reducing the solution mixture with Zn+H_(2)SO_(4) is :
Answer» `4.64 mL` `5.46 mL` `3.46 mL` `2.64 mL` SOLUTION :MEQ. Of `Kr_(2)Cr_(2)O_(7)=Meq. Of Fe^(2+)` [from `Fe+` from `Fe_(2)(SO_(4))_(3)`] `V=4.64 mL`

Discussion

92757.	1g pure iron is dissolved in excess of H_(2)SO_(4). The clear filtrate is made up 100 mL. 10mL of this solution is treated with 0.1 M KMnO_(4) solution till whole of the Fe^(2+) ions are oxidised to Fe^(3+) ions. Now 0.2 g Fe_(2)(SO_(4))_(3) is dissolved in it. the solution is now treated with Zn and H_(2)SO_(4). The volume of 0.1 M KMnO_(4) used after reducing the solution mixture with Zn+H_(2)SO_(4) is:
Answer» `5.572 mL` `3.572 mL` `4.572 mL` `6.572 mL` SOLUTION :MEQ. of `FE^(2+)` [FORM `Fe`]+Meq. of `Fe^(2+)` [from `Fe_(2)(SO_(4))_(3))`]=Meq of `KMnO_(4)` `1/56xx1000xx10/100+0.2/(400/2)xx1000=0.1xx5xxV` `:. V=([1.786+1])/0.5=5.572 mL`

Discussion

92758.	1g pure iron is dissolved in excess of H_(2)SO_(4). The clear filtrate is made up 100 mL. 10mL of this solution is treated with 0.1 M KMnO_(4) solution till whole of the Fe^(2+) ions are oxidised to Fe^(3+) ions. Now 0.2 g Fe_(2)(SO_(4))_(3) is dissolved in it. the solution is now treated with Zn and H_(2)SO_(4). The amount of K_(2)Cr_(2)O_(7) to be dissolved to prepare V mL of K_(2)Cr_(2)O_(7), which is just sufficient to completely oxidised 10 mL of above FeSO_(4) solution ?
Answer» `0.0875 G` `0.875 g` `8.75 g` `0.0087g` Solution :Meq. Of `Kr_(2)Cr_(2)O_(7)` needed for `FeSO_(4)` =meq of `Fe` in `10 mL` `w/49xx1000/VxxV=1/56xx1000x10/100` `:. W=0.0875 g`

Discussion

92759.	1g pure iron is dissolved in excess of H_(2)SO_(4). The clear filtrate is made up 100 mL. 10mL of this solution is treated with 0.1 M KMnO_(4) solution till whole of the Fe^(2+) ions are oxidised to Fe^(3+) ions. Now 0.2 g Fe_(2)(SO_(4))_(3) is dissolved in it. the solution is now treated with Zn and H_(2)SO_(4). The volume of KMnO_(4) needed to convert Fe^(2+) ions to Fe^(3+) ions in 100 mL original solution is:
Answer» `71 ML` `142 mL` `35.7 mL` `80 mL` Solution :Meq. Of `Fe^(2+)`=Meq. Of `KMnO_(4)` (for `100 mL` solution) `1/56xx1000=0.1xx5xxV` `:. V=35.7 mL`

Discussion

92760.	1g of NaCl is dissolved in 10g of a solution the density of which is 1.07g//"cc". Find the molality and molarity of NaCl.
Answer» Solution :(i) LET us consider one litre of sodium thiosulphate solution `:.` WT of the solution `=` DENSITY `xx` volume (mL) `=1.25xx1000=1250g` Wt. of `Na_(2)S_(2)O_(3)` present in 1L of the solution `=` molarity `xx` mol.wt. `=3xx158=474g` Wt `%` of `Na_(2)S_(2)O_(3)=(474)/(1250)xx100=37.92%` (ii) Wt. of solute `(Na_(2)S_(2)O_(3)=474g` Moles of solute `=(474)/(158)=3` Wt. of solvent `(H_(2)O)=1250-474=776g` Moles of solvent `=(776)/(18)=43.11` `:.` mole fraction of `Na_(2)S_(2)O_(3)=(3)/(3+43.11)=0.063` (iii) MOLALITY of `Na_(2)S_(2)O_(3)=("moles of " Na_(2)S_(2)O_(3))/("wt.of solvent in grams")xx1000` `=(3)/(776)xx1000=3.865m` `:.1` mole of `Na_(2)S_(2)O_(3)` contains 2 moles of `NA^(+)` ions and 1 mole of `S_(2)O_(3)^(2-)` ions `:.` molality of `Na^(+)=2xx3.865=7.73m` Molality of `S_(2)O_(3)^(2-)=3.865m`

Discussion

92761.	1g of Mg is burnt with 0.28g of O_(2) in a closed vessel. Which reactant is left in excess and how much?
Answer» Mg,5.8g Mg,0.58g `O_(2),0.24g` `O_(2),2.4g` SOLUTION :`2MG(s)+O_(2)(g)rarr2MgO` 48g 32G 80g 32g `O_(2)-=48gMg` `0.28g O_(2)-=(48)/(32)xx0.28g Mg` =0.42g Mg Excess `Mg=1-0.42=0.58g`

Discussion

92762.	1g of a monabasic acid when dissolved in 100g of water lowers the freezing point by 0.168^(@)C. 0.2g of the same acid when dissolved and titrated required 15.1mL of N//10 alkali . Calculate the degree of dissociation of the acid. (K_(f) for water is 1.86)
Answer» Solution :Milliequivalent of alkali (m.e.) `=(1)/(10)xx15.1=1.51` (Eqn. 1, Chapter 7) m.e. of the ACID `=1.51` (Eqn. 2 , Chapter 7) `:.` eq. of acid `=(1.51)/(1000)=0.00151` (Eqn.3, Chapter 7) Now equivalent of acid `=("weight in grams")/("equivalent weight")` `=(0.2)/("eq.wt.")` `=(0.2)/(M)` [for monobasic acid , eq. wt. =MOL. wt. (M)] `:.(0.2)/(M)=0.00151`, `M=132.45` `:.` MOLALITY of the acid (m)= `("moles of solute")/("wt. of solvent in grams")xx1000` `=(1)/(132.45)xx(1000)/(100)=0.076` We have, `(DeltaT_(f))_("normal")=K_(f).m` `=1.86xx0.076=0.141` `:.i=((DeltaT_(f))_("observed"))/((DeltaT_(f))_("normal"))`.......(Eqn 9) `=(0.168)/(0.141)=1.19` The monobasic acid (say AH) ionises as moles before diss : `{:(1"mole",,,,0,,,,0),(AH,,=,,A^(-),,+,,H^(+)):}` moles after diss : `{:((1-x),,x,,x,,(x="degree of dissociation")):}` `i=(1-x+x+x)/(1)=1.19` ..............(Eqn 10) `:.x=0.19`

Discussion

92763.	1F equals to ______.
Answer» 96500 moles 96500 C `1.6 TIMES 10^(-19)C` `1.6 times 10^(-19)` moles Answer :B

Discussion

92764.	1=Butyne on reaction with hot alkalineKMnO_4 gives:
Answer» `CH_3 CH_2 CH_2COOH` `CH_3 CH_2 COOH + CO_2` `CH_3 CH_2 COOH` `CH_3COOH + HCOOH ` ANSWER :B

Discussion

92765.	19g fused SnCl_(2) was electrolysed using inert electrodes. 0.119g Sn was deposited at cathode. If nothing was given out during electrolysis, calculate the ratio of weight of SnCl_(2) and SnCl_(4) in fused state after electrolysis. ("At. Wt. "Sn = 119)
Answer» ANSWER :`71.34 ;`

Discussion

92766.	19.7kg of gold was recovered from a smulggler. The atoms of gold recovered are: (Au=197)
Answer» 100 `6.02xx10^(23)` `6.02xx10^(24)` `6.02xx10^(25)` ANSWER :D

Discussion

92767.	19.7 kg of gold was recovered from a smuggler. How many atoms of gold were recovered ? (Au = 197).
Answer» 100 `6.02 XX 10^23` `6.02 xx 10^24` `6.02 xx 10^25` Solution :197 G of gold ` = 6.02 xx 10^23 ` atoms` 19.7 xx 10^3 g` of gold `(6.02 xx 10^23)/(197) xx 19.7 xx 10^3= 6.02 xx 10^25`

Discussion

92768.	19.6 gm of ferrous ammonium sulphate [FeSO_4(NH_4)_2SO_4 .6H_2O] were dissolved and made up to 500 ml with acidified water. 25 ml of this solution required 20 ml and 27.5 ml of the solution A and B of KMnO_4 respectively. How many ml of A must be added to 1 litre of B to make N/10 KMnO_4 solution.
Answer» ANSWER :370

Discussion

92769.	19.5 g of CH_(2)FCOOH is dissolved in 500 g of water. The depression in the freezing point observed is 1.0^(@)C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid. K_(f) for water is "1.86 K kg mol"^(-1).
Answer» Solution :Here, `w_(2)=19.5g, w_(1)=500g, K_(f)="1.86 K kg mol"^(-1),(DeltaT_(f))_("obs")=1.0^(@)` `therefore""M_(2)"(observed)"=(1000K_(f)w_(2))/(w_(1)DeltaT_(f))=(("1000 g kg"^(-1))("1.86 K kg mol"^(-1))("19.5 g"))/("(500 g)""(1.0 K)")="72.54 g mol"^(-1)` `M_(2)"(calculated) for "CH_(2)FCOOH=14+19+45="78 g mol"^(-1)` `"van't Hoff FACTOR (i)"=((M_(2))_("cal"))/((M_(2))_("obs"))=(78)/(72.54)=1.0753.` Calculation of dissociation constant. Suppose degree of dissociation at the given concentration is `ALPHA`. `{:("Then",CH_(2)FCOOH,hArr,CH_(2)FCOO^(-),+,H^(+),),("Initial","C mol L"^(-1),,,,,),("At EQM.",C(1-alpha),,Calpha,,Calpha",","Total "=C(1+alpha)):}` `therefore""i=(C(1+alpha))/(C)=1+alpha"or"alpha=i-1=1.0753-1=0.0753` `K_(a)=([CH_(2)FCOO^(-)][H^(+)])/([CH_(2)FCOOH])=(Calpha.Calpha)/(C(1-alpha))=(Calpha^(2))/(1-alpha)` Taking volume of the solution as 500 mL, `C=(19.5)/(78)xx(1)/(500)xx1000=0.5M""therefore""K_(a)=(Calpha^(2))/(1-alpha)=((0.5)(0.0753)^(2))/(1-0.0753)=3.07xx10^(-3)`.

Discussion

92770.	19.5 g of CH_(2)FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0^(@)C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.
Answer» Solution :`w_(1)=500 g "" w_(2)=19.5 g` `K_(f)=1.86" K kg mol"^(-1) "" Delta T_(f)=1 K` We known that : `M_(2)=(K_(f)XX w_(B)xx 1000)/(Delta K_(f)xx w_(1))` `= (1.86" K kg mol"^(-1)xx19.5 g xx 1000 g kg^(-1))/(500 g xx 1K)` `= 72.54 mol^(-1)` Therefore, observed molar mass of `CH_(2)FCOOH, (M_(2))_(obs)=72.54` g mol The calculated mass of `CH_(2)FCOOH` is : `(M_(2))_(cal)=14+19+12+16+16+1` `= 76 mol^(-1)` Therefore, van.t HOFF factor, `i=((M_(2))_(cal))/((M_(2))_(obs))` `= (78 g mol^(-1))/(72.54 g mol^(-1))=1.0753` Let ABE the degree of dissociation of `{:("At",CH_(2)FCOOH,hArr,CH_(2)FCOO^(-),+,H^(+)),("equilibrium","C mol L"^(-1),,O,,O),(,C(1-alpha),,C alpha,,C alpha):}` Total `= C(1-alpha)` `i=1+alpha` `alpha =i-1=1.0753-1` = 0.0753 Now, the value of `K_(a)` is given as : `K_(a)=([CH_(2)FCOO^(-)][H^(+)])/([CH_(2)FCOO])` `=(C alpha C alpha)/(C(1-alpha))=(C alpha^(2))/(1-alpha)` Taking the volume of the solution as 500 mL, we have the concentration : `C=((19.5)/(78))/(500)xx1000 M` =0.5 M Therefore, `K_(a)=(C alpha^(2))/(1-alpha)` `= (0.5(0.0753)^(2))/(1-0.0753)` `= (0.5xx0.00567)/(0.9247)` = 0.00307 (approximately) `= 3.07xx10^(-3)`

Discussion

92771.	1.94 g of a mixture of KOH (56) and K_(2)CO_(3)(138) is dissolved in water and separated into two equal parts by volume. One part required 50 mL 0.1 M H_(2)SO_(4) to reach the phenolphthalein end point while the other part required 75 mL of the same acid to reach the methyl orange end point. The mass percentage of K_(2)CO_(3) in the mixture is
Answer» `35.5%` `71%` `29%` `64.5%` Solution :Let one PART of solution has x mmol KOH and y mmol `K_(2)CO_(3)`. `implies x + y = 10` and x + 2y = 15 `implies y = 5, x = 5` Total mmol of `K_(2)CO_(3)` in the sample = 10 MASS of `K_(2)CO_(3)=(10 XX 138)/(1000)= 1.38 g` `implies` Mass % of `K_(2)CO_(3)= (1.38)/(1.94) xx 100 = 71%`

Discussion

92772.	19 g of molten SnCl_(2) is electrolysed for same tine, using inert electrodes. 0.119 g of Sn is deposited at the cathode. No substance is lost during the electrolysis. Find the ratio of the weight of SnCl_(2) and SnCl_(2) after electrolysis (Sn = 119)
Answer» Solution :According to this problem `SnCl_(2)` should first ionise into the IONS of tin and chloride. Sn gets depositedat the cathode and as `Cl_(2)`, PRODUCED at anode, is not lost, it combines with the remaining `SnCl_(2)` to give `SnCl_(4)`. `SnCl_(2) rarr Sn + Cl_(2)` `SnCl_(2) + Cl_(2)rarr SnCl_(4)` Equivalent of Sn deposited `= (0.119)/(119//2) = 0.002 ""("eq. wt. of Sn " = (119)/(2))` `therefore` equivalent of `SnCl_(2)` decompose = 0.002. Equivalent of CHLORINE produced = 0.002. Equivalent of `SnCl_(4)` = formed = 0.002. `therefore` weight of `SnCl_(4)` formed = no. of eq. `xx` eq. wt. `= 0.002 xx (261)/(2) = 0.261` gram. `(("eq. wt. of " SnCl_(4) = (261)/(2)" , in " SnCl_(2) + Cl_(2), rarr SnCl_(4)),("+2",+4))` Further eq. of chlorine produced = initial wt. of `SnCl_(2)` - (wt. of `SnCl_(2)` decomposed+ wt. of `SnCl_(2)` combined with `Cl_(2)` to form `SnCl_(4)`) = 19 - (eq. of `SnCl_(2)` decomposed `xx` eq. wt. of `SnCl_(2)` + eq. of `SnCl_(2)` which combined with `Cl_(2) xx` eq. wt. of `SnCl_(2)`). `= 19 - [0.002 xx (190)/(2) + 0.002 xx (190)/(2)]` `= 19 - 0.38` `= 18.62g`. `therefore SnCl_(2) : SnCl_(4) = 18.62 : 0.261`. (weight ratio)

Discussion

92773.	1.9 8m of a sample of H_(2)O_(2) solution containing y% H_(2)O_(2) be weight requires y ml of KMnO_(4) solution for complete titration under acidic condition. Find the molarity of KMnO_(4) solution
Answer» 0.2 M 0.11 M 0.011 M 0.25 M Answer :B

Discussion

92774.	19 g molten SnCl_(2) was electrolysed by two inert electrodes. On electrolysis, 0.119g of Sn was deposited at cathode. Find the ratio of weights of SnCl_(2) and SnCl_(4) after electrolysis, assuming, there is no loss of materíal during electrolysis.
Answer» Solution :`Sn^(2+)+2e rarrSn,` electricity required to produce `"0.11 g Sn "=(2xx96500)/(119)xx0.199=193C` `2SnCl_(2)rarrSn+SnCl_(4)or, 2Sn^(2+)rarr Sn+Sn^(4+)` `Sn^(2+)rarrSn^(4+)+2e" HENCE, "2xx96500C "-="1 mol "Sn^(4+)` `therefore""193C-=10^(-3)" mol "Sn^(4+)` `therefore` AMOUNT of `SnCl_(4)` PRODUCED `=10^(-3)xx(119+142)=0.261g` `"0.119 g Sn "=10^(-3)" mol Sn "-=10^(-3)" mol "SnCl_(2) and 10^(-3)" mol "SnCl_(4)-=10^(-3)" mol "SnCl_(2)` `therefore` To produce `10^(-3)` mol Sn and `10^(-3)` mol `SnCl_(2)2xx10^(-3)` mol `SnCl_(2)` is required. `therefore""SnCl_(2)` LEFT after electrolysis `=(19-2xx10^(-3)xx190)=18.62g, SnCl_(4)` produce `=10^(-3)xx(119+4xx35.5)=0.261g.` `therefore"Ratio of the AMOUNTS of "SnCl_(2)" to "SnCl_(4)=(18.62)/(0.261)=71.34`

Discussion

92775.	18g of glucose is dissolved in 1000g of water at 300K. At what temperature does this solution boil?(Kb for water is 0.52 K kg/mol.Molar mass of glucose is 180 g/mol, boiling point of water = 273.15 K)
Answer» Solution :`"Given 18 g of SOLUTE "=OMEGAT""DeltaT_(b)=K_(b)xxm` `"Molecular mass "C_(6)H_(12)O_(6)=180""DeltaT_(b)=0.52xx0.1` `omegat"of the solvent 1000g"DeltaT_(b)=0.052K` `"B.P of Water"=373.15K` `"w.K.t"DeltaT_(b)=T_(b)-T_(b)^(@)` `0.052=T_(b)-373.15` `T_(b)=373.15+0.052` `"What will boil at "T_(b)=373.202K`

Discussion

92776.	18gm water is added in a 200 gm sample of Oleum labeled as 109%. The new labelling of the final sample is :
Answer» `106%` `103%` `100%` `0%` Solution :MINIMUM labelling of OLEUM `= 100%` All free `SO_(3)` is exhausted using `18GM` of water and pure `218gm H_(2)SO_(4)` formed.

Discussion

92777.	18g of glucose (C_(6)H_(12)O_(6)) is added to 178.2g of water. The vapour pressure of water (in torr.) for this aqueous solution is
Answer» `76.0` `752.4` `759.0` `7.6` ANSWER :B

Discussion

92778.	18g each of A: maltose B : Sucrose C: Lactose A are separately taken and hydrolysed comple tely. The order of increase in the mass of products formed with each of them.
Answer» `A GT B gt C` `B gt A gt C` `C gt A gt A` `A=B=C` ANSWER :D

Discussion

92779.	""^(18)F undergoes 90% decay in 365 min. The decay constant for ""^(18)F is :
Answer» `6.31xx10^(-3)MIN""^(-1)` `7.32xx10^(-3)min""^(-1)` `6.31xx10^(-4)min""^(-1)` `6.31xx10^(-2)min` Solution :`lamda=(2.303)/(365)LOG""(100)/(10)` `=(2.303)/(366)xx1=6.31xx10^(-3)min""^(-1)`

Discussion

92780.	18.97 g fused SnCl_(20 was electrolyzed using inert electrodes. 1.187g Sn was deposited at cathode. If nothing is obtained during electrolysis, calculate the ration of weight of SnCl_(2) and SnCl_(4) in fused state after electrolysis. Given: Atomic weight of Sn=118.7, Mw of SnCl_(2)=189.7, Mw of SnCl_(4)=260.7
Answer» Solution :Fused `SnCl_(2)overset(El ectrolysis)rarr SN^(2+)+2Cl^(c-)` At cathode `: Sn^(2+)+2e^(-)rarr Sn,,,(` Reduction `)` At anode `: 2Cl^(c-)rarr Cl_(2)(g)+2e^(-),,,(` Oxidation `)` Further , `Cl_(2)(g)` formed at anode REACTS with left over `SnCl_(2)` to give `SnCl_(4)` . `SnCl_(2) +Cl_(2) rarr SnCl_(4)` During ELECTROLYSIS`:` Eq of `SncL_(2)` lost `=` Eq of `Cl_(2)` formed `=` Eq of `Sn` formed. `Eq` of `Sn` formed `=(W_(Sn))/(Ew of Sn)` `=(1.187)/(118.7//2) ""Sn^(2+)+2e^(-)rarrSn` =`2xx10^(-2)""n` FACTOR of `Sn=2` `Eq` of `SnCl_(2)` lost `=Eq` of `Cl_(2)` formed `=Eq `of `SnCl_(4)` formed `=Eq` of `Sn` formed `=2xx10^(-2)` Now total loss is equivalent of `SnCl_(2)` during complete course `=Eq` of `SnCl_(2)` lost during electrolysis is `+ Eq` of `SnCl_(2)` lost during reaction with `Cl_(2)` `=2xx10^(-2)+2xx10^(-2)=4xx10^(-2)` `Eq` of `SnCl(` initially `)=(18.97g)/(189.7//2)=2xx10^(-1)` `Eq` of `SnCl_(2)` left is molten solution `=2xx10^(-1)-4xx10^(-2)=0.16` `Eq` of `SnCl_(4)` formed `=2xx10^(-2)=0.02` `("WEIGHT of SnCl left")/("Weight of SnCl formed ")=("Weight of "SnCl_(2)"left" xx Ew of SnCl_(2))/("Weight of "SnCl_(4) "formed "xx Ew of SnCl_(4))` `=((0.16xx189.7//2))/((0.02xx260.7//2))=5.82`

Discussion

92781.	._(18)Ar^(40), ._(20)Ca^(40) and ._(19)K^(40) are
Answer» Isomers Isotopes Isobars Isotones Solution :Atoms of DIFFERENT ELEMENTS having different atomic no.but same mass no. are called isobars

Discussion

92782.	1.84 g of mixture ofCaCO_3 and MgCO_3 are heated strongly till no further loss of weight takes place. The residueweighs 0.96g Find the percentage composition of the mixture.
Answer» ANSWER :`CaCO_3`= `54.35%` and `MgCO_3= 45.65%`

Discussion

92783.	1.84 g of a mixture of CaCO_(3) and MgCO_(3) is strongly heated till no further loss of mass takes place. The residue weighs 0.96 g. Calculate the percentage composition of the mixture.
Answer» Solution :Suppose the mass of `CaCO_(3)` in the mixture = x g. `THEREFORE` Mass of `MgCO_(3)` in the mixture will be = (1.84 - x) g. Step 1. To calculate the mass of CAO RESIDUE from x g calcium carbonate. `underset(100 g)(CaCO_(3))overset(Delta)rarr underset(56 g)(CaO)+CO_(2)` Thus, 100 g of `CaCO_(3)` upon decomposition give a residue of CaO = 56 g `therefore` x g of `CaCO_(3)` will give residue of `CaO=(56xx x)/(100)=0.56xx x g` Thus, the mass of CaO residue formed = 0.56 x g Step 2. To calculate the mass of MGO residue from (1.84 - x) of magnesium carbonate. `underset(84 g)(MgCO_(3))overset(Delta)rarr underset(40 g)(MgO+CO_(2)` 84 g of `MgCO_(3)` upon decomposition yield residue of MgO = 40 g `therefore (1.84 - x)g` of `MgCO_(3)` will yield residue of `MgO=(40xx(1.84-x))/(84)g` Thus, the mass of MgO residue formed `=(40)/(84)(1.84-x)g`. Step 3. To calculate the masses of `CaCO_(3)` and `MgCO_(3)` in the mixture. Total mass of CaO and MgO residue from Step 1 and Step 2 `=0.56 x + (40)/(84)(1.84 - x)=0.96 g` (Given) or `0.56 x xx 84+40xx1.84-40x=0.96xx84`or `47.04x -40x -40x=80.64-73.60` or `7.04x=7.04`or x = 1 Thus, the mass of `CaCO_(3)` in the mixture = 1.00 g and the mass of `MgCO_(3)` in the mixture `= 1.84-1.00 = 0.84 g`. % of `CaCO_(3)` in the mixture `=(1)/(1.84)xx100=54.35` % of `MgCO_(3)` in the mixture = 100 - 54.35 = 45.65.

Discussion

92784.	18.4 A current is passed for 1 hour and 42 minutes through CuSO_(4) solution having graphite electrode at 298K temperature and 1 bar pressure. If cell has capacity of 75% then calculate the mass of Cu and volume of O_(2). [Cu=63.5u,O=16u,R=0.08314]
Answer» ANSWER :`37.05g" "CU and 5.63L " "O_(2)`

Discussion

92785.	1.80 g of a certain metal burnt in oxygen gave 3.0 g of its oxide. 1.50 g of the same metal heated in steam gave 2.50 g of its oxide. Show that these results illustrate the law of constant proportion.
Answer» Solution :In the first sample of the OXIDE, WT. of metal=1.80 G, wt. of oxygen =(3.0-1.80)g =1.2 g `therefore("wt. of metal")/("wt. of oxygen")=(1.80 g)/(1.2 g)=1.5` In the SECOND sample of the oxide, wt of metal =1.50 g, wt. of oxygen=(2.50-1.50)g =1 g `therefore ("wt. of metal")/("wt. of oxygen")=(1.50g)/(1 g)=1.5` Thus, in both samples of the oxide the proportions of the weights of the metal and oxygen are fixed. Hence, the results follows the law of constant proportion. Note : This law is not applicable in CASE of isotopes.

Discussion

92786.	180 electron rule: Valence shell of a transition metal consists of nine valence orbitals of (n-1) d, ns and np sub-shells, which can accommodate 18 electrons. When a metal complex has 18 valence electrons, it is said to have achieved the same electronic cofiguration as the noble gas at the end of the period. The rule and its exception are similar to the application of the octet rule of main group elements. The rule is not helpful for complexes of the s-block metals, the lanthanides or actinides. Effective atomic number is the total number of electrons possesed by the central possesed by the central metal atom/ion including the electrons contributed by the ligands. It is obvious that if a central transition metal atom/ion achieves 18 electrons in its valence shell, it attains the nearest noble gas cofiguration, which is known as effective atomic numbr (EAN) rule. [Atomic number :Cr=24, Fe=26, Cu=29, Ni=28, Rh=45] Which of the following does not obey EAN rule or 18 electron rule ?
Answer» `[CR(CO)_(6)]` `[Fe(Ox)_(3)]^(4-)` `[CU(CN)_(4)]^(3-)` `[Ni(NH_(3))_(6)]^(2+)` SOLUTION :`[Cr(Co)_(6)] to 24+6xx2=36` `[Fe(Ox)_(3)]^(-4) to 26-2+4xx3=36` `[Cu(CN)_(4)]^(3-) to29-1+4xx2=36` `[Ni(NH_(3))_(6)]^(2+) to 28-2+6xx2=38`

Discussion

92787.	180 electron rule: Valence shell of a transition metal consists of nine valence orbitals of (n-1) d, ns and np sub-shells, which can accommodate 18 electrons. When a metal complex has 18 valence electrons, it is said to have achieved the same electronic cofiguration as the noble gas at the end of the period. The rule and its exception are similar to the application of the octet rule of main group elements. The rule is not helpful for complexes of the s-block metals, the lanthanides or actinides. Effective atomic number is the total number of electrons possesed by the central possesed by the central metal atom/ion including the electrons contributed by the ligands. It is obvious that if a central transition metal atom/ion achieves 18 electrons in its valence shell, it attains the nearest noble gas cofiguration, which is known as effective atomic numbr (EAN) rule. [Atomic number :Cr=24, Fe=26, Cu=29, Ni=28, Rh=45] 18 electron rule is applicable to :
Answer» s-block ELEMENTS p-block elements d-block elements f-block elements Answer :C,

Discussion

92788.	18 ml of mixture of acetic acid and sodium acetate required 6 ml of 0.1 M NaOH ForNeutrilisation of the acid and 12 ml of 0.1 M HCl for reaction with salt Seperately . If Pk_a of the acid is 4.75 What is the pH pf the mixture.
Answer» 5.05 4.75 4.5 4.6 Answer :A

Discussion

92789.	18 mL of mixture of acetic acid and sodium acetate required 6 mL of 0.1 M NaOH for neutralization of the acid and 12 mL of 0.1 M HCI for reaction with salt, separately. If pK_(a) of the acid is 4.75, what is the pH of the mixture ? [log 2=0.3]
Answer» `5.05` `4.75` `4.5` `4.6` ANSWER :a

Discussion

92790.	18 ml o 1.0 M Br_(2) solution undergoes complete disproportionation in basic medium to Br_(ϴ) and Br_(3)^(ϴ) . Then the resulting solution requires 45 ml of As^(+3) solutioni to reduce BrO_(3)^(ϴ) to Br^(ϴ).As^(+3) is oxidised toAs^(+5) which statements are correct?
Answer» `E_(w)(Br_(2))=M/10` `E_(w)(Br_(2))=(5M)/3` Molarity of `As^(+3)=0.4M` Molarity of `As^(+3)=0.2M` ANSWER :B::C::D

Discussion

92791.	18g glucose (C_(6)H_(12)O_(6)) is added to 178.2g water. The vapour pressure of water (in torr) for this aqueous solution is:
Answer» 759 739.6 `746.0` 752.4 Answer :D

Discussion

92792.	18 g of water is taken to prepare, the tea. Find out the internal energy of vaporisation at 100^(@)C. (Delta_(vap)H^(ө) for water at 373K=40.66kJmol^(-1))
Answer» 37.56 kJ `mol^(-1)` `-37.56" kJ "mol^(-1)` `43.76" kJ "mol^(-1)` `-43.76" kJ "mol^(-1)` Solution :`underset((18g))(H_(2)O(l)) to underset((18g))(H_(2)O(G))` Number of moles in 18 g of `H_(2)O(l)=(18g)/(18g" "mol^(-1))=1` mol `Delta_(vap)U=Delta_(vap)H^(ө)-Deltan_(g)RT` `=40.66-(1)(8.314)(373)J" "mol^(-1)` `=40.66-3.10=37.56` kJ `mol^(-1)`

Discussion

92793.	18 g of water contains :
Answer» 1 g ATOM of hydrogen 2 g ATOMS of hydrogen 3g atoms of hydrogen None of the above Answer :B

Discussion

92794.	1.8 g of metal were deposited by a current of 3 ampere for 50 minute. The equivlent wt. of metal is :
Answer» `20.5` `25.8` `19.3` `30.7` ANSWER :C

Discussion

92795.	1.8 g of glucose is dissolved in 100 g of water in a beaker. At what temperature will the solution boil if the pressure is 1.013 bar ? Given that the boiling point of pure water at 1.013 ba is 373.15 K and K_(b) for is 0.052K kg mol^(-1).
Answer» Solution :`M_(B)=(K_(b)xxW_(B))/(DeltaT_(b)xxW_(A)) or DeltaT_(b)=(K_(b)xxW_(b))/(M_(B)xxW_(A))` `W_(B)=1.8g, W_(A)=100g=0.1" Kg", M_(B)=180" g mol"^(-1), K_(b)=0.052" K kg mol"^(-1)` `DeltaT_(b)=((0.052" K kg mol"^(-1))xx(1.8 g))/((180" g mol"^(-1))xx(0.1" kg"))=0.005 K` Boiling POINT of pure water = 373.15 K Boiling point of solution = (373.15+0.005)K =373.155 K

Discussion

92796.	18 g of glucose, C_(6)H_(12)O_(6) (Molar Maas=180 g mol^(1))is dissolved in 1 kg of water in a sauce pan. Al what temperature will this solution boil ? (K_(b) for water = 0.52 K kg mol^(-1), boiling point of pure water = 373.15 K )
Answer» Solution :`Delta T_(B) = K_(b) xx m` `T_(b) - T_(b)^(0) = 0.52 K " KG " mol^(-1)` `xx (18 g)/(180 " g " mol^(-1) ) xx (1)/(1kg) ` `T_(b) - 372.15 = (0.52)/(10)` `T_(b) - 373.15 = 0.052 `K `T_(b) = 0.052 + 373.15` `T_(b) = 373.202` K

Discussion

92797.	18 g of glucose, C_(6)H_(12)O_(6) is dissolved in 1 kg of water in a saucepan. At what temperature wil the water boil at 1.013 bar pressure ? K_(b) for water is 0.52 K kg mol^(-1).
Answer» Solution :Here, we are given `w_(2)=18 G, w_(1)=1 kg = 1000 g, K_(B)=0.52 "kg mol"^(-1)` `M_(2) ("for glucose", C_(6)H_(12)O_(6)) = 72+12+96=180 g mol^(-1)` `Delta T_(b)=(1000 K_(b)w_(2))/(w_(1)M_(2))=(1000" g kg"^(-1)xx0.52"K kg mol"^(-1)xx18 g)/(1000 g xx 180 " g mol"^(-1))=0.52 K` As water BOILS at 37.3 15 K at 1.013 bat PRESSURE, therefore, boiling point of solution `= 373.15+0.052 K = 373.202 K`.

Discussion

92798.	18 g of glucose (C_(6)H_(12)O_(6)) is dissolved in 1 kg of water in a saucepn. At what temperature will water boil under 1.013 bar pressure ? Given K_(b) for water is 0.52 K kg mol^(-1)
Answer» Solution :`W_(B)=(M_(B)xxDeltaT_(b)xxW_(A))/K_(b) or DeltaT_(b)=(W_(B)xxK_(b))/(M_(B)xxW_(A)` Mass of solute `(W_(B))`=18g Mass of WATER `(W_(B))`=1 kg Molar mass of solute `(M_(B))`= 0.52 K kg `mol^(-1)` Molal elevation constant `(K_(b))`=0.52 K kg `mol^(-1)` `DeltaT_(b)=((18g)XX(0.52 K kg mol^(-1)))/((180 g mol^(-1))xx(1 kg))=0.052 K` At 1.013 bar PRESSURE (atmospheric pressure), BOILING point of water= 37300K Boiling point of solution= `(373.0+0.052 K)`

Discussion

92799.	18 g of glucose (C_(6)H_(12)O_(6)) is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100^(@)C is :
Answer» 7.60 torr 76.00 torr 752.40 torr 759.00 torr Solution :`(p^(@)-p)/(p^(@))=x`(solute) `p^(@)`, vapour PRESSURE of pure WATER at `100^(@)C` `=760` torr `x=((18)/(180))/((18)/(180)+(1782)/(18))=(01)/(01+99)=(01)/(10)` `:.(01)/(10)=(760-p)/(760)` Solving `p=752*4` torr.

Discussion

92800.	18 g of glucose (C_(6) H_(12) O_(6)) (molar mass = 180 g mol^(-1)) is dissolved in 1 kg of water in a sauce pan . At what temperature will this solution boil ? (Kb for water = 0.52 K kg mol^(-1) , boiling point of water = 373.1 K)
Answer» SOLUTION :`373.202` K

Discussion

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