94420 + Interview Questions in Chemistry in Class 12 Page 1743 InterviewSolution

87101.	A solution of Sodium metal in liquid ammonia is strongly reducing agent due to a
Answer» Sodium ATOMS Sodium hydride SODAMIDE Solvated ELCTRONS

Discussion

87102.	A solution of sodium chloride in water is electrolysed using inert electrodes . The solution is formed in vessel is
Answer» `NaOH` `H_(2)O` `NaCl` `HCl` Solution :`Na^(+) + OH^(-) to NaOH`

Discussion

87103.	A solution of sodium bicarbonate in water turns
Answer» Phenolphthalein pink Methyl ORANGE yellow Methyl orange red Blue litmus red Solution :`NaHCO_(3)` in WATER is alkaline in nature due to HYDROLYSIS of `HCO_(3)^(-)` ION. `NaCO_(3) hArr Na^(+) + HCO_(3)^(-)`.

Discussion

87104.	A solution of sodium chloride in contact with atmosphere has a pH of about
Answer» 3.5 5 7 1.4 Solution :When strong acid and strong BASE are react NEUTRAL SALT are formed. So that NaCl is a neutral salt.

Discussion

87105.	A solution of SO_2 in water reacts with H_2S precipitating sulphur. Here SO_2 acts as:
Answer» An OXIDISING agent A REDUCING agent An acid A catalyst Answer :A

Discussion

87106.	A solution of SO_2 in water reacts with H_2S precipitating sulphur. Here SO_2 acts as
Answer» an OXIDISING agent a REDUCING agent an ACID a catalyst SOLUTION :`2H_2S + SO_2 to 3S + 2H_2O`

Discussion

87107.	A solution of silver nitrate is electrolysed for 20 minutes with a current of 2 ameres. Calculate the mass of silver deposited at the cathode.
Answer» Solution :Electrochemical reaction at cathode is `AG^(+) + e^(-) to Ag` (reduction) `m = ZI t` `m - (108 g mol^(-1))/(96500 C mol^(-1))xx 2400 C` `m = 2.68 g` `Z = ("molarmass of Ag")/((96500)) = (108)/(1 xx 96500)` `I = 2A` `t = 20 xx 60S = 1200S` `It = 2A xx 1200 S = 2400C`.

Discussion

87108.	A solution of silver benzoate has a pH of 8.63. K_a (C_6H_5COOH) = 6.5 xx 10^(-5), Calculate the value of K_(sp)for silver benzoate.
Answer» Solution :USE PH ` = 1/2 (pk_w + pk_a+ log C)` `1.4 xx 10^(-2)`

Discussion

87109.	A solution of salt in HCl when diluted with excess of water turns milky. It indicates the presence of
Answer» `AL` `BI` `Sb` `Zn` ANSWER :B::C

Discussion

87110.	A solution of salt in HCl when diluted with water turns milky. It indicates the presence of:
Answer» Sn Bi Sb All Answer :D

Discussion

87111.	A solution of protein (extracted from crabs) was prepared by dissolving 0.75g in 125 ml of aqueous solution. At 4^(@) C, an osmotic rise of the solution was observed. Then the molecular weight of protein is (assume density of solution is 1.0 g/ml)
Answer» `9.4 XX 10^(5)` `5.4 xx 10^(5)` `5.4 xx 10^(5)` `9.5 xx 10^(10)` ANSWER :C

Discussion

87112.	A solution of protein (extracted from crabs) was prepared by dissolving 0.75 g in 125 cm^(3) of an aqueous solution. At 4^(@)C,an osmotic pressure rise of 2.6 mm of the solution was observed. Thenmolecular weight of protein is (Assume density of solution is 1.00 g//cm^(3))
Answer» `9.xx10^(5)` `5.4xx10^(5)` `5.4xx10^(10)` `9.4xx10^(10)` Solution :Given h = 2.4 mm = 0.26 CM, V = 125 `cm^(3)`

Discussion

87113.	A solution of protein (extracted from crabs) was prepared by dissolving 0.75 g in 125 cm^3 of an aqueous solution. At 4^@C an osmotic pressure rise of 2.6 mm of the solution was observed. Then molecular weight of protein is (Assume density of solution is 1.00 g/cm^3):
Answer» `9.4xx10^5` `5.4xx10^5` `5.4xx10^10` `9.4xx10^10` ANSWER :B

Discussion

87114.	A solution of potassium chromate is treated with an excess of dilute nitric acid. Then the observations is
Answer» `CR^(3+)` and `Cr_20_7^(2-)` are formed `Cr_20_7^(2-)` and `H_2O` are formed `Cr_20_7^(2-)` is REDUCED to +3 state of Cr `Cr_20_7^(2-)` is oxidised to +7 state of Cr Answer :B

Discussion

87115.	A solution of pH=5. it is diluted 100 times, then it will become……
Answer» NEUTRAL basic unaffected more acidic Solution :pH =5 MEANS `[H^+]=10^-5` After dilution `[H^+]=10^-5//100=10^-7M` `[H^+] from H_2O` cannot be NEGLECTED. TOTAL `[H^+]=10^-7+ 10^-7=2 times 10^-7` `pH=7-0.3010=6.6990=7` `pH=7` (neutral)

Discussion

87116.	A solution of ph 2.0 is more acidic than the one with ph 6.0 by a factor of
Answer» 3 4 3000 10000 Answer :D

Discussion

87117.	A solution of Ni(NO_(3))_(2) is electrolysed between platinum electrodes using a current of 5.0 ampere for 20 minutes. What mass of nickel will be deposited at the cathode? (At. Mass of Ni=58.7)
Answer» Solution :Quantity of electricity PASSED=(5A)`xx(20xx60s)=6000C` `NI^(2+)+2e^(-)toNi` Thus, `2F`, i.e., `2xx96500C` deposit Ni=1 MOLE, i.e., 58.7g `therefore6000C` will deposit `Ni=(58.7)/(2xx96500)xx6000g=1.825g`.

Discussion

87118.	A solution of Pb(NO_(3))_(2) is added dropwise to a second solution in which [Cl^(-)]=[F^(-)]=[I^(-)]=[SO_(4)^(2-)]=0.001 M. What is the first precipitate that forms?
Answer» `PbCl_(2)""(K_(sp)=1.5xx10^(-5))` `PbF_(2)""(K_(sp)=3.7xx10^(-8))` `PbI_(2)""(K_(sp)=8.5xx10^(-9))` `PbSO_(4)""(K_(sp)=1.8xx10^(-8))` Answer :d

Discussion

87119.	A solution of Ni(NO_(3))_(2)is electrolyzed between platium electrodes using a current of 5A for 20 mi n. What mass of Ni is deposited at the cathode ?
Answer» SOLUTION :`W=zxxixxt` `Z=(58.7)/(2xx96500)` `w="1.825 GRAMS"`

Discussion

87120.	A solution of Ni(NO_3)_2 is electrolysed between platinum electrodes using acurrent of 5 amperes for 20 minutes. What mass of nickel is deposited at thecathode?[molar mass of Ni = 58.7gma " mol"^(-1)]
Answer» Solution :Therefore MASS DEPOSITED by `5 xx 20 xx 60 C = (58.7 xxx 6000)/(2 xx 96500) = 1.84 g` or `W = Zit` `w = (29.35 xx 5 xx 20 xx 60)/(96500) = 1.83 g` RELATION used : Mass of nickel deposited `(W) = (I t E)/(96500) g`. where Current (i)= 5 amperes time (t) = 20 minutes `= 20 xx 60 = 1200 s` Equivalent mass (E) `= ("Molar mass")/("total (+)ve charge") = (58.7)/(2)` `=29.35 g mol^(-1)` or, `W = ("it E")/(96500)` `= (5 xx 1200 xx 29.35)/(96500)` Mass of Nickel deposited = 1.8248 g.

Discussion

87121.	A solution of Ni(NO_(3))_(2) is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode ?
Answer» Solution :* Mass of electricity `Q=5Axx20xx60s` =6000C * By the following reaction NI is formed. Ionization : `Ni(NO_(3))_(2(aq)) to Ni_((aq))^(2+)+2NO_(3(aq))^(-)` REDUCTION on cathode: `Ni_((aq))^(2+) + underset("2 mol "e^(-))(2e^(-)) to underset("1 mol Ni")(Ni_((l)))` * According to this reaction, 2 mol `e^(-)`, 2F electricity 1 mol `Ni=58.7` g Ni So, `2xx96500` COULOMB electricity produce 58.79 gm Ni and so mass of Ni obtained by using 6000 coulomb electricity `=(6000xx58.79)/(2xx96500)` `=1.8277g` Ni

Discussion

87122.	A solution of Ni(NO_3)_2 is electrolysed between platinum electrodes using a current of 0.5 ampere for 20 minutes. What mass of Ni is deposited at the cathode ?
Answer» Solution :Quantity of electricity passed `=(5A) xx (20 xx 60 s) = 6000 C` `Ni^(2+) + 2e^(-) to Ni` Thus, 2F, i.e. `2 xx 96500 C` will DEPOSIT = 1mole, i.e. 58.7 g of Ni `therefore 6000 C` will deposit `=58.7/(2 xx 96500 ) xx 6000 g = 1.825 g` of Ni.

Discussion

87123.	A solution of [Ni(H_(2)O_(6)]^(2+) is green but a solution of [Ni(CN)_(4)]^(2-)is colourless. Explain.
Answer» Solution :In `[NI(H_(2)O_(2))_(6)]^(2+)`. Ni is in +2 STATE with the configuration `3d^(8)` i.e., it has two unpaired electrons which do not pair up in the presence of the weak `H_(2)O` LIGAND. Hence, it is COLOURED. The d-d transition, absorbs red light and the complementary light emitted is green In CASE of `[Ni(CN)_(4)]^(2-)` Ni is again in +2 state with the configuration `3d^(@)` but in presence of the strong CN ligand, the two unpaired electrons in the 3d orbitals pair up. Thus, there is no unpaired electron present. Hence, it is colourless

Discussion

87124.	A solution of [Ni(H_(2)O_(6)]^(2+) is green, whereas a solution of [Ni(CN)_(4)]^(2-) is colourless-Explain.
Answer» Solution :`(i)` In `[Ni(H_(2)O)_(6)]^(2+)` , `Ni` is in `+2` oxidation state with the configuration `3d^(8)`.i.e., it has two unpaired electrons which do not pair up in the presence of WEAK `H_(2)O` ligand. HENCE, it is coloured. The `d-d` transtion ABSORBS red light and the complementary light emitted is green. `(ii)` In the case of `[Ni(CN)_(4)]^(2-)`, `Ni` is again in `+2` oxidation state with the configuration `3d^(8)`, but in the presence of strong `CN^(-)` ligand the two unpaired electrons in the `3d` orbitals pair up. THUS there is no unpaired electron present. Hence it is colourless. Therefore, a solution of `[Ni(H_(2)O)_(6)]^(2+)` is green, whereas a solution of `[Ni(CN)_(4)]^(2-)` is colourless.

Discussion

87125.	A solution of [Ni(H_(2)O)_(6)]^(2+) is green but a solution of [Ni(CN)_(4)]^(2-) is colourless. Explain.
Answer» Solution :In `[NI(H_(2)O)_(6)]^(2+)`, Ni is in +2 state with the configuration `3d^(8)`, i.e., it has TWO unpaired electrons which do not pair up in the presence of the weak `H_(2)O` LIGAND. Hence, it is coloured. The d-d transition absorbs red LIGHT and the complementary light emitted is green In CASE of `[Ni(CN)_(4)]^(2-)`, Ni is again in +2 state with the configuration `3d^(8)` but in presence of the strong `CN^(-)` ligand, the two unpaired electrons in the 3d orbitals pair up. Thus, there is no unpaired electron present. Hence, it is colourless.

Discussion

87126.	A solution of [Ni(H_2 O)_6]^(2+) is green but a solution of [Ni(CN)_4]^(2-) is colourless . Explain.
Answer» Solution :In `[NI(H_2 O)_6]^(2-)` Ni is in +2 state with the CONFIGURATION `2d^8`,i.e., it has two unpaired electrons which do not pair up in the presence of the weak `H_2 O` ligand. So, it is coloured. The d-d transition, ABSORBS red light and the complementary light emitted is green. In case of `[Ni(CN)_4]^(2-)` , Ni is again in +2 state with the configuration `3d^8` but in presence of the STRONG `CN^-` ligand, the two unpaired electrons in the 3d-orbitals pair up. Hence, there is no unpaired ELECTRON present, Hence, it is colourless.

Discussion

87127.	A solution of Ni (NO_3)_2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode.
Answer» Solution :Quantity of electricity PASSED =5(A)` times 20 times 60 (S)` =6000C Electrode reaction is `Ni^(2+)+2E^(-) to Ni` `2 times 96,500C` of electricity product Ni `=(58.7 times 6000)/(2 times 96500)` =1.825 G

Discussion

87128.	A solution of NaCl in contact with atmosphere has a pH of about
Answer» `3.5` 5 7 `14` SOLUTION :NaCIis a SALTOS strongacid andstrongbasethereforeits SOLUTIONWILL beneutrali.e.,pH=7

Discussion

87129.	A solution of Na_2SO_4in water is electrolysed using inert electrodes. The products at cathode and anode are respectively
Answer» `O_2 , H_2` `O_2 , NA` `H_2, O_2 ` `O_2, SO_2` ANSWER :C

Discussion

87130.	A solution of Na_(2)S_(2)O_(3) is standardised iodometrically against 3.34 g of pure KBrO_(3) (converted to Br^(-)), requiring 40 mL Na_(2)S_(2)O_(3) solution. What is the molarity of Na_(2)S_(2)O_(3) solution ? (Molar mass of KBrO_(3)=167 g "mol"^(-1))
Answer» Solution :Necessary equations : `KBrO_(3)+KIrarrI_(2)Br^(-)` `I_(2)+Na_(2)S_(2)O_(3)rarrNa_(2)S_(4)O_(6)+`NAL equivalent of `I_(2)` = equivalent of `KBrO_(3)=12XX10^(-2)` equivalent of `Na_(2)S_(2)O_(3)=`equivalent of `KBrO_(3)=12xx10^(-2)` equivalent of `Na_(2)S_(2)O_(3)`=equivalent of `I_(2)` `Mxx1xx(40)/(1000)=12xx10^(-2)` so MOLARITY = 3M.

Discussion

87131.	(A) Solution of Na_2CrO_4 in water is intensely coloured. (R) Oxidation state of Cr in Na_(2)CrO_(4) is +VI
Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) Both (A) and (R) are true and (R) is not the correct explanation of (A) (A) is true but (R) is FALSE Both (A) and (R) are false Solution :Both (A) and (R) are true and (R) is not the correct explanation of (A)

Discussion

87132.	A solution of Na_2CO_3that is 2.0 M in CO_3^(2-)ions was boiled with excess of CaF_2 . Very small amounts of CaCO_3and F^-were formed. If the solubility product of CaCO^3is x and molar solubility of CaF_2is y, find the molar concentration of F^-in the resulting solution after equilibrium is attained.
Answer» SOLUTION :`SQRT((8y^3)/(X))`

Discussion

87133.	A solution of monobasic acid with molarity 3xx10^(-2)M has a freezing point depression of 0.06^(@)C. Calculate pK_(a) of the acid (Molal depression constant of water is 1.86^(@)C//m)
Answer» Solution :`DeltaT_(f)=I XX K_(f)xxm` Taking molality = MOLARITY (as the solution is DILUTE) `0.06=i xx 1.86xx3xx10^(-2)"or"i=(0.06)/(1.86xx3xx10^(-2))=1.0753` For the acid HA, `{:(,HA ,hArr,H^(+),+,A^(-),),("Initial moles",C,,0,,0,),("At. eqm.",C(1-alpha),,Calpha,,Calpha,"Total "=C(1-alpha)+Calpha+Calpha=C(1+alpha)):}` `therefore"i"=("Total no. moles after dissociation")/("Inital moles")=(C(1+alpha))/(C)=1+alpha` `"or"alpha=i-1=1.0753-1=0.0753` `K_(a)=(Calpha.Calpha)/(C(1-alpha))=Calpha^(2)` `=(3xx10^(-2))(0.0753)^(2)=1.74xx10^(-4)""(therefore alpha lt lt 1)` `pK_(a)=-logK_(a)=-log(1.74xx10^(-4))=3.769~=3.77`

Discussion

87134.	A solution of MgCl_(2) in water has pH
Answer» `lt 7` `GT 7` 7 14.2 Solution :`MgCl_(2)+ 2H_(2)O hArr Mg(OH)_(2) + 2HCL`

Discussion

87135.	A solution of metal salt was electrolysed for 15 minutes with a current of 1.5 A. The mass of the metal deposited was 0.000783 kg. Calculate the equivalent mass of the metal.
Answer» Solution :`Q=1.5xx900C=1350C.` 96500 C DEPOSITE 1 g eq. HENCE, `E=(0.783)/(1350)xx96500=55.97`

Discussion

87136.	A solution of metal ion when treated with Kl gives a red precipitate which dissolves in excess KI to give a colourless solution. Moreover, the solution of metal ion on treatment with a solution of cobalt(II) thiocyanate gives rise to a deep blue crystalline precipitate. The metal ion is
Answer» `Pb^(2+)` `Hg^(2+)` `Cu^(2+)` `Co^(2+)` SOLUTION :`Hg^(2+)+"KL"toHgL_(2)("Red ppt.")` `Hgl_(2)+underset(EXCESS)(Kl)tounderset("Soluble")(K_(2)Hgl_(4))` `Hgl^(2+)+Co(SCN)_(2)tounderset("blue CRYSTALLINE precipiates")(Hg(SCN)_(2))` Hence,(B) is the correct answer.

Discussion

87137.	A solution of methylene blue is passed through animal charcoal. The filtrate obtained will appear
Answer» BLUE green COLOURLESS RED SOLUTION :colourless

Discussion

87138.	A solution of KOH hydrolysis CH_(3)CHClCH_(2)CH_(3) and CH_(3)CH_(2)CH_(2)CH_(2)Cl. Which one of these is more easily hydrolysed ?
Answer» Solution :Under aqueous conditions, substitution OCCURS by `S_(N)1` mechanism since carbocation intermediates are stabilized by solvation. Further, the reactivity in `S_(N)1` reactions depends upon the STABILITY of CARBOCATIONS. `CH_(3)-CHCl-CH_(2)CH_(3) overset("Ionization")to underset(2^(@)" carbocation (more stable)")(CH_(3)-overset(+)(C)H-CH_(2)CH_(3)+Cl^(-))` `CH_(3)CH_(2)CH_(2)CH_(2)-Cl overset("Ionization")to underset(1^(@)" carbocation (less stable)")(CH_(3)CH_(2)CH_(2)overset(+)(C)H_(2)+Cl^(-))` Now since `CH_(3)CHClCH_(2)CH_(3)` upon ionization gives the more stable `2^(@)` carbocation while `CH_(3)CH_(2)CH_(2)CH_(2)Cl` on ionization gives the less stable `1^(@)` carbocation, therefore, `CH_(3)CHClCH_(2)CH_(3)` hydrolyses FASTER than `CH_(3)CH_(2)CH_(2)CH_(2)Cl`.

Discussion

87139.	A solution of lead nitrate prepared by dissolving 2.07 g ofpure lead innitricacid was treated with HCI, Cl_(2) gas and NH_(4) Cl . Whatwill be the maximumweightof (NH_(4))_(2) Pb Cl_(6) so produced ?
Answer» Solution :We have, `underset(2.07 g ) (Pb) to Pb(NO_(3))_(2) underset (Cl_(2)NH_(4)Cl)overset(HCI)to (NH_(4))_(2) Pb Cl_(6) ` Now , for MAXIMUM yield of `(NH_(4))_(2) PbCl_(6)` molesof Pb in the reactant = moles of Pb in `(NH_(4))_(2) Pb Cl_(6)` ` 1 xx ` moles of `(NH_(4))_(2) Pb Cl_(6)` or `(2 . 07)/( 207) = ("maximum wt. of"(NH_(4))_(2) Pb Cl_(6))/(456)""[(NH_(4))_(2) Pb Cl_(6) = 456]` `:.` maximumwt. of `(NH_(4))_(2)PbCl_(6) = 4 . 56 g`

Discussion

87140.	A solution of KOH hydrolyses CH_(3)CHClCH_(2)CH_(3) " and " CH_(3)CH_(2)CH_(2)CH_(2)Cl . Which one of these is more easily hydrolysed ?
Answer» Solution :Assuming `S_(N)2` reaction , `CH_(3)CH_(2)CH_(2)CH_(2)CL ` will be more easily hydrolysed on STERIC consideration. `CH_(3)CHClCH_(2)CH_(3)` will face some steric hindrance.

Discussion

87141.	A solution of KOH in water is called :
Answer» POTASH lye soda lye Salt cake None of these Answer :A

Discussion

87142.	A solution of KMnO_(4) on reduction yields either a colourless solution or a brown precipitate or a green solution depending on pH of the solution. What different stages of the reduction do these represent and how are they carried out?
Answer» Solution :OXIDISING behaviour of `KMnO_(4)` depends on pH of the solution. In acidic medium `(pH lt 7)`. `MnO_(4)^(-) + 8H^(+) + 5e^(-) rarr underset(("Colourless"))(MN^(2+)) + 4H_(2)O` In alkaline medium `(pH gt 7)`, `MnO_(4)^(-) + E^(-) rarr underset(("Green"))(MnO_(4)^(2-))`

Discussion

87143.	A solution of KMnO_(4) on reduction yields either a colourelesssolution or a brown precipitate or a green solution depending on Ph of the solution. What different stages of the reduction do these represent and how are they carried out ?
Answer» Solution :`KMnO_(4)` is anoxidizing agent. Its OXIDISING behaviour depends UPON pH of the solution. In acidic medium `(pH LT 7) , MnO_(4)^(-) + 8H^(+) + 5e^(-) rarr underset("(Coloureless)")(Mn^(2+)) + 4H_(2)O` In alkaline medium `( pHgt 7) , MnO_(4)^(-) + e^(-) rarr underset("(Green)")(MnO_()^(2-))("manganese")` Inneutral medium `( pH= 7 ) , MnO_(4)^(-)+ 2H_(2)O + 3E^(-) rarr underset("(Brown PPT.)")(MnO_(2)) + 4OH^(-)`

Discussion

87144.	A solution of KBr will liberate Br_(2)with
Answer» `I_(2)` `Cl_(2)` `SO_(2)` `HI` ANSWER :B

Discussion

87145.	A solution of impure KMnO_4 is prepared by dissolving 2g of it in IL of the solution. 20 ml of this solution required 24.2 ml of N/(20) sodium oxalate solution. The percentage purity of KMnO_4 sample is :
Answer» `95.6%` `92.2%` `98.0%` `84%` Solution :`N_1V_1-=N_2V_2` `N_1 XX 20 = (N)/(20) xx 24.2` `N_1= (24.2)/(400) =(6.05)/(100)` Strength `=(6.05)/(100) xx 31.6 = 1.91` g % Purity `=(1.91)/(2.0) xx 100 = 95.5%`

Discussion

87146.	A solution of Hg^(2+) ion ontreatment with asolutionof cobalt (II) throcyanategives rise to adeep bluecrystallineprecipitate .Then thecoordinationnumberof mercuryinthe deepbluecolouredcompound is
Answer» Solution : `NH_(6)^(Theta) + underset("DEEP blue ppt cobaltcorathocyanate macomnite")[CO(NO_(2))_(2) RARR Co[HG(SCN)_(6)]`

Discussion

87147.	A solution of HCI is prepared by dissolving 5.5 g HCI in 200 g ehanol. The density of the solution is 0.79 g mL^(-1) Molarity of solution is :
Answer» 0.58 M 0.16 M 0.92 M 1.2 M SOLUTION :`"Molarity (M)"=("Mass of HCI//Molar mass")/("Volame of solution in litrese")` Mass of HCI solution = 200+5.5=205.5 g DENSITY of solution = 0.79 g `mL^(-1)` `"VOLUME of solution"=((205.5g))/((0.79gmL^(-1)))` =260L `"Molarity"=((5.5g)//(36.5g mol^(-1)))/(0.260L)` =0.58 `mL^(-1)=0.58 M`

Discussion

87148.	A solution of glycerol (C_(3)H_(8)O_(3)), molar mass ="92 g mol"^(-1), in water was preapred by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42^(@)C. What mass of glycerol was dissolved to make this solution? K_(b) for water ="0.512 K kg mol"^(-1).
Answer» SOLUTION :`DeltaT_(B)=(100K_(b)w_(2))/(w_(1)M_(2))"or"w_(2)=(w_(1)M_(2)DeltaT_(b))/(1000K_(b))=(500xx92xx0.42)/(1000xx0.512)=37.7g`

Discussion

87149.	A solution of glycerol (C-(3)H_(8)O_(3) , molar mass = 92 g mol^(-1) iin water was prepared by dissolving some glycero 500 g of water. This solution has a boiling point of 100.42^(@)C. What mass of glycerol was dissolved to make this solution ? K_(b) for water= 0.512 k kg mol^(-1).
Answer» Solution :`DeltaT_(B)=(100.42^(@)C-100)^(@)C=0.42^(@)C=0.42" K",` `M_(B)=92" G mol"^(-1),W_(A)=500 g =0.5" kg",` `K_(b)=0.152" k kg mol"^(-1)` `W_(B)=(M_(B)xxDeltaT_(b)xxW_(A))/(K_(b))=((92g mol^(-1))xx(0.42K)xx(0.5kg))/((0.512" k kg mol"^(-1)))=37.7 g`.

Discussion

87150.	A solution of glycerol (C_(3) H_(8) O_(3)) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42^(@)C while pure water boils at 100^(@)C. What mass of glycerol was dissolved to make the solution ? (K_(b) = 0.512 K kg mol^(-1))
Answer» SOLUTION :37.73 G

Discussion

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A solution of Pb(NO_(3))_(2) is added dropwise to a second solution in which [Cl^(-)]=[F^(-)]=[I^(-)]=[SO_(4)^(2-)]=0.001 M. What is the first precipitate that forms?

A solution of Ni(NO_(3))_(2)is electrolyzed between platium electrodes using a current of 5A for 20 mi n. What mass of Ni is deposited at the cathode ?

A solution of Ni(NO_3)_2 is electrolysed between platinum electrodes using acurrent of 5 amperes for 20 minutes. What mass of nickel is deposited at thecathode?[molar mass of Ni = 58.7gma " mol"^(-1)]

A solution of Ni(NO_(3))_(2) is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode ?

A solution of Ni(NO_3)_2 is electrolysed between platinum electrodes using a current of 0.5 ampere for 20 minutes. What mass of Ni is deposited at the cathode ?

A solution of [Ni(H_(2)O_(6)]^(2+) is green but a solution of [Ni(CN)_(4)]^(2-)is colourless. Explain.

A solution of [Ni(H_(2)O_(6)]^(2+) is green, whereas a solution of [Ni(CN)_(4)]^(2-) is colourless-Explain.

A solution of [Ni(H_(2)O)_(6)]^(2+) is green but a solution of [Ni(CN)_(4)]^(2-) is colourless. Explain.

A solution of [Ni(H_2 O)_6]^(2+) is green but a solution of [Ni(CN)_4]^(2-) is colourless . Explain.

A solution of Ni (NO_3)_2 is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode.

A solution of NaCl in contact with atmosphere has a pH of about

A solution of Na_2SO_4in water is electrolysed using inert electrodes. The products at cathode and anode are respectively

A solution of Na_(2)S_(2)O_(3) is standardised iodometrically against 3.34 g of pure KBrO_(3) (converted to Br^(-)), requiring 40 mL Na_(2)S_(2)O_(3) solution. What is the molarity of Na_(2)S_(2)O_(3) solution ? (Molar mass of KBrO_(3)=167 g "mol"^(-1))

(A) Solution of Na_2CrO_4 in water is intensely coloured. (R) Oxidation state of Cr in Na_(2)CrO_(4) is +VI

A solution of monobasic acid with molarity 3xx10^(-2)M has a freezing point depression of 0.06^(@)C. Calculate pK_(a) of the acid (Molal depression constant of water is 1.86^(@)C//m)

A solution of MgCl_(2) in water has pH

A solution of metal salt was electrolysed for 15 minutes with a current of 1.5 A. The mass of the metal deposited was 0.000783 kg. Calculate the equivalent mass of the metal.

A solution of metal ion when treated with Kl gives a red precipitate which dissolves in excess KI to give a colourless solution. Moreover, the solution of metal ion on treatment with a solution of cobalt(II) thiocyanate gives rise to a deep blue crystalline precipitate. The metal ion is

A solution of methylene blue is passed through animal charcoal. The filtrate obtained will appear

A solution of KOH hydrolysis CH_(3)CHClCH_(2)CH_(3) and CH_(3)CH_(2)CH_(2)CH_(2)Cl. Which one of these is more easily hydrolysed ?

A solution of lead nitrate prepared by dissolving 2.07 g ofpure lead innitricacid was treated with HCI, Cl_(2) gas and NH_(4) Cl . Whatwill be the maximumweightof (NH_(4))_(2) Pb Cl_(6) so produced ?

A solution of KOH hydrolyses CH_(3)CHClCH_(2)CH_(3) " and " CH_(3)CH_(2)CH_(2)CH_(2)Cl . Which one of these is more easily hydrolysed ?

A solution of KOH in water is called :

A solution of KMnO_(4) on reduction yields either a colourless solution or a brown precipitate or a green solution depending on pH of the solution. What different stages of the reduction do these represent and how are they carried out?

A solution of KMnO_(4) on reduction yields either a colourelesssolution or a brown precipitate or a green solution depending on Ph of the solution. What different stages of the reduction do these represent and how are they carried out ?

A solution of KBr will liberate Br_(2)with

A solution of impure KMnO_4 is prepared by dissolving 2g of it in IL of the solution. 20 ml of this solution required 24.2 ml of N/(20) sodium oxalate solution. The percentage purity of KMnO_4 sample is :

A solution of Hg^(2+) ion ontreatment with asolutionof cobalt (II) throcyanategives rise to adeep bluecrystallineprecipitate .Then thecoordinationnumberof mercuryinthe deepbluecolouredcompound is

A solution of HCI is prepared by dissolving 5.5 g HCI in 200 g ehanol. The density of the solution is 0.79 g mL^(-1) Molarity of solution is :

A solution of glycerol (C_(3)H_(8)O_(3)), molar mass ="92 g mol"^(-1), in water was preapred by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42^(@)C. What mass of glycerol was dissolved to make this solution? K_(b) for water ="0.512 K kg mol"^(-1).

A solution of glycerol (C-(3)H_(8)O_(3) , molar mass = 92 g mol^(-1) iin water was prepared by dissolving some glycero 500 g of water. This solution has a boiling point of 100.42^(@)C. What mass of glycerol was dissolved to make this solution ? K_(b) for water= 0.512 k kg mol^(-1).

A solution of glycerol (C_(3) H_(8) O_(3)) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42^(@)C while pure water boils at 100^(@)C. What mass of glycerol was dissolved to make the solution ? (K_(b) = 0.512 K kg mol^(-1))