94420 + Interview Questions in Chemistry in Class 12

1.	Concentration of reaction and products at equilibrium for A + 2B hArr C + D are, [A] = 0.20, [B] =0.10, [C] = 0.30, [D] = 0.50. The value of equilibrium constant is:
Answer» 75 150 2.5 750 Answer :A

Discussion

2.	Concentration of CN in 0.1 M HCN is [K_(a) = 4 xx 10^(-10)]
Answer» `2.5 xx 10^(-6)M` `4.5 xx 10^(-6)M` `6.3 xx 10^(-6)M` `9.2 xx 10^(-6)M` SOLUTION :CONCENTRATION of `CN^(-) = sqrt(K_(a)C) = 6.3 xx 10^(-6)`.

Discussion

3.	Concentration of copper glance is done by
Answer» Leaching Magnetic separation Hydraulic washing Froth Floation method Answer :D

Discussion

4.	Concentration of Ca ion in blood plasma is(approximately)
Answer» `10 mg L^(-1)` `100 mg L^(-1)` `50 mg L^(-1)` `25 mg L^(-1)` ANSWER :B

Discussion

5.	Concentration of Cl^(-) ions in a solution obtained by mixing 600 ml of 0.1 M CaXl_(2) and 400 ml of 0.4 M AlCl_(3).
Answer» `0.28 M` `0.6 M` `0.22 M` `0.54 M` SOLUTION :`{:(CaCl_(2),rarr,Ca^(+2),+,2Cl^(-),),(0.1 xx (600)/(1000),,,,0.12 "mole",),(AlCl_(3),,rarr,AL^(+3) + 3 Cl^(-),,),(0.4 xx (400)/(1000) " mole",,,,0.48 " mole",):}` `n_(Cl^(-)) = (0.12 + 0.48) " mole" = 0.6 " mole"` `V_(T) = V_(1) + V_(2) = (400 + 600) mL` = 1 litre `M_(Cl) = (0.6)/(1) = 0.6 M`

Discussion

6.	Concetration of a reactant A is changed from 0.044 M to 0.032 M in 25 minutes, the average rate of the reactio during this interval is
Answer» `0.0048"mole/lit/min"` `0.00048"mole/lit/sec"` `4.8xx10^(-4)"mole/lit/min"` `0.0048"mole/lit/sec"` ANSWER :C

Discussion

7.	Concentration described by percentage is commonly used in industrial, chemical industries.
Answer» ANSWER :TRUE

Discussion

8.	Concentration is expressed in ?
Answer» `("number of MOLES / LITRE")/("TIME in SEC")` `("time in sec")/("number of moles / litre")` `("number of moles / litre")/("volume")` `("volume")/("number of moles / litre")` ANSWER :A

Discussion

9.	concentrated sulphuric acid on reaction with NaCl, NaBr and Nal produces HCl, bromine and iodine respectively. What order of oxidising ability of halogens with reference to sulphuric acid can be established on the basis of this reaction ?
Answer» `H_(2)SO_(4) GT l_(2) gt Br_(2) gt Cl_(2)`<BR>`Cl_(2) gt H_(2)SO_(4) gt Br_(2) gt l_(2)` `H_(2)SO_(4)gt Cl_(2) gt Br_(2) gt l_(2)`s `Cl_(2) gt Br_(2) gt l_(2) gt H_(2)SO_(4)` ANSWER :B

Discussion

10.	Concentrated sulphuric acid is approximately 18 molar. 5 cc of it are added to make 500 cc of the solution. The approxi mately normality of the solution will be
Answer» 0.18 0.09 0.36 0.27 Solution :`18 M H_(2)SO_(4) EQUIV 36 N H_(2)SO_(4)` `:. 5 xx 36 = 500 xx N_(2) "or" N_(2) = 0.36`

Discussion

11.	Concentrated sulphuric acid is added followed by heating in each of the following test types labelled (i) to (v) Identify in which of the above test tubes, the following changes will be observed. Support your answer with the help of a chemical equation. disappearance of yellow powder along with the evolution of a colourless gas.
Answer» SOLUTION :`S_(8)+16H_(2)SO_(4)rarr24SO_(2)+16H_(2)O`

Discussion

12.	Concentrated sulphuric acid is added followed by heating of each of the following test tubes labelled (i) to Which of the following statement is incorrect about these observations ?
Answer» Evolution of reddish brown GAS take place in (ii) test tube and black substance is formed in (i) test tube Evolution of reddish brown gas take place in (iii) test tube and blue solution is formed Evolution of neutral COLOURLESS gas take place in (iv) test tube and gas burns with blue flame Evolution of YELLOWISH green gas take place in (v) test tube and colourless solution is formed Solution :(i)`C_12H_22O_11underset("DEHYDRATION")overset(Conc. H_2SO_4)12C (black)+11H_2O` (ii)`2NaBr+2H_2SO_4 to Na_2SO_4 + SO_2+Br_2 uarr` (reddish brown) +`2H_2O` (iii)`2Cu(NO_3)_2+2H_2SO_4to2CuSO_4` (blue solution) +`4NO_2 uarr` (reddish brown) +`O_2+2H_2O` (iv)`K_4[FE(CN)_6]+6H_2SO_4+6H_2Oto2K_2SO_4+FeSO_4+3(NH_4)_2SO_4+6COuarr`(neutral and colourless ) Co burns with blue flame. (v)`KCl+H_2SO_4 to KHSO_4 + HCl uarr` (colourless)

Discussion

13.	Concentrated sulphuric acid is added followed by heating in each of the following test types labelled (i) to (v) Identify in which of the above test tubes, the following changes will be observed. Support your answer with the help of a chemical equation. formation of brown substance which on dilution becomes blue.
Answer» SOLUTION :`Cu+2H_(2)SO_(4)rarrCuSO_(4)+SO_(2)+2H_(2)O`

Discussion

14.	Concentrated sulphuric acid is added followed by heating in each of the following test types labelled (i) to (v) Identify in which of the above test tubes, the following changes will be observed. Support your answer with the help of a chemical equation. evolution of colourless gas
Answer» SOLUTION :`KCl+H_(2)SO_(4)OVERSET(Delta)rarrKHSO_(4)+HCl(G)`

Discussion

15.	Concentrated sulphuric acid is added followed by heating in each of the following test types labelled (i) to (v) Identify in which of the above test tubes, the following changes will be observed. Support your answer with the help of a chemical equation. formation ofblack substance
Answer» Solution :`C_(12)H_(22)O_(11)OVERSET("Conc."H_(2)SO_(4))rarrunderset(("BLACK mass"))(12C+11H_(2)O)`

Discussion

16.	Concentrated sulphuric acid is added followed by heating in each of the following test types labelled (i) to (v) Identify in which of the above test tubes, the following changes will be observed. Support your answer with the help of a chemical equation. evolution of brown gas
Answer» Solution :`2NaBr+3H_(2)SO_(4)OVERSET("heat")rarr2NaHSO_(4)+SO_(2)+Br_(2)+2H_(2)O`

Discussion

17.	Concentrated sulphuric acid is added followed by heating in each of the following test tubes labelled (i) to (v) : Identify in which of the above test tubes, the following changes will be observed. Support your answer with the help of a chemical equation. (a) formation of black substance (b) evolution of brown gas (c ) evolution of colourless gas (d) formation of brown subtance which on dilution becomes blue. (e ) disappearance of yellow powder along with the evolution of a colourless gas.
Answer» SOLUTION :(a) Black substance will be formed in tube (i), Actually, concentrated sulphuric ACID is a powerful dehydrating agent. It dehydrates sugar `(C_(12)H_(22)O_(11))` completely to leave a black residue of carbon with the smell of burning sugar. This is called CHARRING of sugar. `underset(("Sugar"))(C_(12)H_(22)O_(11)) overset("Conc."H_(2)SO_(4)) underset("heat") to underset(("Black mass")) (12C+ 11H_(2)O)` (b) Brown gas will evolve in tube (ii). It is bromine formed on heating sodium bromide with concentrated sulphuric acid. The gas is actually yellowish brown in colour. `2NaBr+3H_(2)SO_(4) overset("heat") to2NaHSO_(4)+SO_(2)+Br_(2)+2H_(2)O` (c ) Colourless gas will evolve in tube (v). It is hydrogen CHLORIDE gas formed by heating POTASSIUM chloride with concentrated sulphuric acid. `KCl+H_(2)SO_(4) overset("heat") to KHSO_(4)+HCl(g)` (d) Brown substance cupric oxide (CuO) will be formed in the tube (iii) It will further react with conc. `H_(2)SO_(4)` to form copper sulphate `(CuSO_(4))` to give a blue solution. `Cu+H_(2)SO_(4) to CuO +SO_(2)+H_(2)O` `CuO+H_(2)SO_(4) to CuSO_(4)+H_(2)O` `ul(Cu+2H_(2)SO_(4) to CuSO_(4)+SO_(2)+2H_(2)O)` (e ) Yellow powder of sulphur will disappearin tube (iv) on reacting with concentrated sulphuric acid. Sulphur dioxide, a colourless gas with pungent smell will evolve during the reaction `H_(2)SO_(4)to H_(2)O+SO_(2)+(O)xx16` `(S_(8)16O to 8SO_(2))/(S_(8)+16 H_(2)SO_(4) to 24SO_(2)+16H_(2)O)`

Discussion

18.	Concentrated sulphuric acid does not act as:
Answer» Efflorescent Hygroscopic Oxidising agent Sulphonating agent Answer :A

Discussion

19.	Concentrated sulphuric acid acts as
Answer» DEHYDRATING agent sulphonating agent reducing agent oxidising agent Answer :A::B::D

Discussion

20.	Concentrated nitric acid used in the laboratory work is 68% nitric acid by mass in aquequs solution. What should be molarity of such sample of the acid if the density of solution is 1.504 g mL""^(-1) ?
Answer» Solution : 68%` HNO_3` means that 68 g `HNO_3 ` is PRESENT in 100 g of solution. Moles of `HNO_3 = (68)/(63 ) = 1.08 ` Density of solution = 1.504 g mL`""^( -1)` VOLUME of solution `=(1000 )/( 1.504 )=66.49` Molarity `= (1.08)/(66.49)xx1000`

Discussion

21.	Concentrated nitric acid used in the laboratory work is 68% nitric acid by mass in aqueoussolution. What should be the molarity of such a sample of the acid if the density of solution is 1.504 g mL^(-1)?
Answer» SOLUTION :In 100 g of nitric acid solution, MASS of nitric acid = 68 g Molar mass of HNO3 = 63 g `"mol"^(-1)` `therefore 68 g HNO_3 = 68/63 ` mole = 1.079 mole DENSITY of solution = `1.504 g mL^(-1)` Molarity of the solution = Mass of Solution/Density of solution `= (1.079 )/(0.0665)M = 16.23 M`

Discussion

22.	Concentrated nitric acid used in the laboratory is 68% nitric acid by mass aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is "1.504 g mL"^(-1)?
Answer» Solution :`"68% nitric acid by mass MEANS that"` `"Mass of nitric acid = 68 G,Mass of solution = 100 g"` `"Molar mass of HNO"_(3)="63 g mol"^(-1)` `therefore"68 g HNO"_(3)=(68)/(63)" mole = 1.079 mole,Density of solution = 1.504 g mL"^(-1)` `therefore"VOLUME of solution "=(100)/(1.504)" mL = 66.5 mL = 0.665 L"` `"Molarity of the solution "=("Moles of the solute")/("Volume of solution in L")=(1.079)/(0.0665)M=16.23M`

Discussion

23.	Concentrated nitric acid used in laboratory work is 68% nitric add by aqueous solution. What should be the molarity of such a sample of the add lf the density of the solution is 1.504 g mL^(-1) ?
Answer» Solution :MASS of NITRIC acid= 68 g Mass of solution = 100 g No. of moles of `HNO_(3) = ("mass ")/("Molar mass")` `= (68)/(63) =1.079` mol Density of solution = 1.504 g `mL^(-2)` Density = `("Mass")/("volume")rArr "volume " = ("mass")/("density")` ` = (100 g)/(1.504 g "mL"^(-1))` = 66.5 mL or 0.0665 L Molarity = `("No.of moles of solute")/("volume of solution in litres ")` = `(1.079 "mol")/(0.09665 L )` = 16.23 mol `L^(-1)` .

Discussion

24.	Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution 1.504 gm L^(-1) is ?
Answer» Solution :Concentrated NITRIC acid used in laboratory work is 68% nitric acid by mass in an aqueous solution. This MEANS that 68 g of nitric acid is dissolved in 100 g of the solution. Molar mass of nitric acid `(HNO_(3))` `= 1xx1+1xx14+3xx16` `= 63 g mol^(-1)` Then, number of moles `HNO_(3)=(68)/(63)mol` = 1.079 mol Given, DENSITY of solution `= 1.504 g mL^(-1)` Therefore, VOLUME of 100 g solution `= (100)/(1.504)mL` `= 66.49 mL` `= 66.49xx10^(-3)L` Molarity of solution `= (1.079 mol)/(66.49xx10^(-3)L)-16.23 M`

Discussion

25.	Concentrated nitric acid used as laboratory reagent is usually 69% by mass of nitric acid. Calculate the volume of the solution which contains "23 g HNO"_(3). (The density of the concentrated nitric acid solution is "1.41 g cm"^(-3)).
Answer» Solution :`"Conc. Of HNO"_(3)="69 G in 100 g of solution = 69 g in 100/1.41 CM"^(3),"i.e" 70.92cm^(3)` `THEREFORE"23 g HNO"_(3)" will be present in solution "=(70.92)/(69)xx23= 23.6cm^(3)`

Discussion

26.	Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation of:
Answer» `NO` `NO_(2)` `N_(2)O` `N_(2)O_(4)` SOLUTION :N//A

Discussion

27.	Concentrated nitric acid reacts with iodine to give:
Answer» HI HOI `HOIO_2` `HOIO_3` ANSWER :C

Discussion

28.	Concentrated nitric acid can be transported in vessels of iron. Why ?
Answer» SOLUTION :Iron METAL is passive to concentrated nitric acid. When Al vessel is FILLED with con. `HNO_3` , a PROTECTIVE OXIDE film of `Fe_2O_3` is formed. This film prevents further reactions of acid with metal. Hence `HNO_(3)` can be transported in iron cans.

Discussion

29.	Concentrated nitric acid is added before proceeding to test for group III members. This is to
Answer» OXIDISE any remaining `H_(2)S` Convert FERROUS ions to ferric ions form nitrates which gives grannular precipitate increases ionisation of ammonium hydroxide Solution :To oxidise ferrous ion into ferrice ion OTHERWISE `FE^(2+)` is not COMPLETELY precipitated as `Fe(OH)_(2)` , as ferrous hydroxide is more soluble in water than `Fe(OH)_(3)`.

Discussion

30.	Concentrated nitric acid on heating decomposes to give:
Answer» `O_2 and N_2` `NO` `O_2` `NO_2 and O_2` ANSWER :D

Discussion

31.	Concentrated nitric acid can be transported in vessels of aluminium or tin cans. Why ?
Answer» Solution :Aluminium and Sn is passive to concentrated nitric ACID. When Al or Sn vessel is FILLED with con. `HNO_(3)`, aprotective oxide FILM of `Al_(2)O_(3)` or `Sn_(2)O_(3)` is formed. This film prevents further reactions of acid with metal. HENCE `HNO_(3)`can be transported in either aluminium or Sn cans.

Discussion

32.	Concentrated hydrochloric acid when kept in open air sometimes produces a cloud of white fumes. The explanation for it is that
Answer» OXYGEN in air reacts with the EMITTED HCI gas to FORM a cloud of chlorine gas STRONG affinity of IICI gas for moisture in air results in forming of droplets of liqiud solutin which APPEARS like a clouds smoke due to strong affinity for atmopheric oxide of sulphur conc. HCI emits strongly smelling HCI gas all the time Solution :HCl gas in pressure of moisture in air forms droplets of liquid solution in the form of cloudy smoke.

Discussion

33.	Concentrated hydrochloric acid when kept in open air sometimes produces a could of white fumes. The explanation for it is that
Answer» Oxygen in air reacts with the emitted HCl gas to form a CLOUD of chlorine gas STRONG affinity of Hcl gas for moisture in air RESULTS in forming of droplets of LIQUID solution which APPEARS like a cloudy smoke . Due to strong affinity for water , concentrated hydrochloric acid pulls moisture of air towards itself . This moisture forms droplets of water . Concentrated hydrochloric acid emits strongly smelling HCl gas all the time Answer :C

Discussion

34.	Concentrated HCl is 10N,1000c c of 1N HCl can be obtained by diluting
Answer» 1 CC of CONC. HCL to 1000 cc 10 cc of conc. HCl to 1000 cc 20 cc of conc. HCl to 1000 cc 100 cc of conc. HCl to 1000 cc Solution :`10xx x=1xx1` `x=0.1` litre=100cc.

Discussion

35.	Concentratged H_(2)SO_(4) is not used to prepare HBr from KBr becaue it
Answer» Oxidizes HBR REDUCES HBr Causes DISPROPORTIONATION of HBr REACTS too SLOWLY with KBr Answer :A

Discussion

36.	Concentrated H_(2)SO_(4)will not dehydrate the following complex :
Answer» `[Cr(H_(2)O)Cl]Cl_(2).H_(2)O` `[Cr(H_(2)O)_(4)Cl_(2)]Cl.2H_(2)O` `[Cr(H_(2)O)_(6)]Cl_(3)` All the above COMPLEXES will be dehydrated. Solution :N//A

Discussion

37.	Concentrated H_(2)SO_(4) is added to each of the five test tubes containing (i) NaBr, (ii) sugar, (iii) sulphur powder, (iv) KCl and (v) copper turnings. The test tubes are then heated. Identify in which of the test tubes the following changes will be observed. The observations are: evolution of a brown gas. Support your answer with the help of a chemical equation in this case.
Answer» SOLUTION :When conc. `H2SO_(4)` reacts with NABR taken in the test tube (i), brown coloured `Br_(2)` gas is evolved. <BR> `2NaBr+3H_(2)SO_(4)overset(DELTA)to2NaHSO_(4)+SO_(2)+Br_(2)uarr+2H_(2)O`

Discussion

38.	Concentrated H_(2)SO_(4) will not dehydrate the following complex:
Answer» `[Cr(H_(2)O)_(5)CL]Cl_(2).H_(2)O` `[Cr(H_(2)O)_(4)Cl_(2)]Cl.2H_(2)O` `[Cr(H_(2)O)_(6)]Cl_(6)` All of these Answer :C

Discussion

39.	Concentrated H_2SO_4 is not used to prepare HBr from KBr because it:
Answer» Oxidizes HBr Reduces HBr Causes DISPROPORTIONATION of HBr Reacts too SLOWLY with HBr Answer :A

Discussion

40.	Concentrated H_2 SO_4 is added to the test tubes containing (a) to (e). {:("Test tube",,"Compounds"),(a.,,"Cane sugar"),(b.,,"sodium bromide"),(c.,,"Copper turnings"),(d.,,"Sulphur powder"),(e.,,"Potassium chloride"):} Identify in which of the above test tubes, the following change will be observed on heating. Also give the chemical equations involved. (i) Formation of black substance. (ii) Evolution of brown gas. (iii) Evolution of colourless gas. (iv) Formation of brown substance, which on dilution becomes blue. (v) Disappearance of yellow powder along with the evolution of a colourless gas.
Answer» Solution :(i) Conc. `H_2 SO_4` + Cane sugar `overset (Delta) rarr` Black substance Conc. `H_2 SO_4` acts a dehydrating AGENT, hence it removes water from cane sugar on heating and forms a black substance KNOWN as sugar charcoal. `C_12 H_22 O_11 overset (Conc. H_2 SO_4)rarr 12 C + 11 H_2 O` Thus, formation of black substance occurs in test + tube `(a)`. (II) Conc. `H_2 SO_4` + sodium bromide `overset (Delta) rarr` Brown gas On heating `NaBr` along with conc. `H_2 SO_4`, brown gas `Br_2` is evolved. Initially `Hbr` is produced, as `HBr` is a moderate reducing agent, it reduces `H_2 SO_4` to `SO_2` and itself is oxidised to bromine gas, `Br_2`. `{:(2NaBr+H_(2)SO_(4) overset(Delta)(rarr)Na_(2)SO_(4)+2HBr),(2HBr +H_(2)SO_(4) overset(Delta)(rarr)Br_(2)+SO_(2)+2H_(2)O),(ulbar(2NaBr +H_(2)SO_(4) rarr Na_(2)SO_(4) +Br_(2) +SO_(2)+2H_(2)O)):}` (iii) Potassium chloride + conc. `H_2 SO_4 overset(Delta) rarr` colourless gas `KCl` on heating with conc. `H_2 SO_4`, liberates `HCl` gas which is colourless. `2KCL + H_2 SO_4(conc.) overset(Delta) rarr K_2 SO_4 + 2 HCl` since `HCl` is not a reducing agent, it does not reduce `H_2 SO_4` to `SO_2`. Thus evolution of colourless gas occurs in test tube `( e)`. (iv) `Cu` turnings + `conc . overset (Delta) rarr` Brown substance `overset (H_2 O) underset ("Blue soln") rarr`. `Cu + 2 H_2 SO_4 overset (Delta) underset ("Anhydrous") CuSO_4 + SO_4 + H_2 O` `CuSO_4 + 5H_2 O overset (Delta) rarr overset ("copper sulphate") underset ("Blue") CuSO_4. 5 H_2 O` On heating, `Cu` turnings with `conc. H_2 SO_4`, a brownish substance, i.e., a mixture of `Cu` turnings and white anhydrous copper sulphate is formed along with evolution of pungent smelling `SO_2` gas. On dilution, the solution TURNS blue due to the formation of hydrated copper sulphate which is soluble in water. Hence, formation of brown substance, which on dilution becomes blue occurs in test tube `(c)`. (v) Sulphur powder + conc. `H_(2)SO_(4) overset(Delta)(rarr)` `{:([H_(2)SO_(4) overset(Delta)(rarr) H_(2)O +SO_(2) +[O]] xx 16),(S_(8)+16[O] rarr 8SO_(2)),(ulbar(underset("Yellow powder")(S_(8)+16H_(2)SO_(4))underset("Colourless pungent smelling gas")(rarr 24SO_(2))+16 H_(2)O)):}` (v) On heating sulphur powder, which is yellowish in colour with conc. `H_2 SO_4`, pungent smelling colourless, `SO_2` gas is produced, this yellow power disappears and the above reaction will occur in test tube (d).

Discussion

41.	Concentrated H_(2)SO_(4) is added to each of the five test tubes containing (i) NaBr, (ii) sugar, (iii) sulphur powder, (iv) KCl and (v) copper turnings. The test tubes are then heated. Identify in which of the test tubes the following changes will be observed. The observations are: evolution of a colourless gas. Support your answer with the help of a chemical equation in this case.
Answer» Solution :When conc. `H_(2)SO_(4)` reacts with KCI taken in the test tube (iv), colourless HCI gas is EVOLVED. `KCl+H_(2)SO_(4)overset(DELTA)toKHSO_(4)+Hcluarr`

Discussion

42.	Concentrated H_2SO_4 has great affinity for:
Answer» `H_2S` `H_2O` `CO_2` `O_2` ANSWER :B

Discussion

43.	Concentrated H_(2) SO_(4) on being heated with zinc yields a gas (A),which when passed through bromine solution produces a white precipitate (D) with BaCI_(2) solution while the acid (C), when heated with concentrated H_(2)SO_(4) yields the original gas (A) and another reddish brown gas (E). Identify B.
Answer» `H_(2) S` `H_(2)SO_(4)` `SO_(2)` `HBr` Solution :`A = SO_2, B = H_2SO_4, C = HBr, D = BaSO_4, E = Br_2` `Zn + 2H_2SO_4 OVERSET(Delta)(RARR)ZnSO_4 + 2H_2O + underset((A))(SO_2)UARR` `Br_2 + 2H_2O + SO_2 rarr underset((B))(H_2SO_4) + underset((C))(2HBr)` `H_2SO_4 + BaCl_2 rarr underset((D))(BaSO_4)darr + 2HCl` `underset((C))(2HBr) + H_2SO_4 overset(Delta)(rarr) 2H_2O + SO_2 + underset((E))(Br_2)`.

Discussion

44.	Concentrated aqueous sulphuric acid is 98% H_(2)SO_(4) by mass and has a density of 1.84"g cm"^(3). What volume of the concentrated acid is required to make 5.0 litre of 0.500 M H_(2)SO_(4) solution?
Answer» Solution :`98% H_(2)SO_(4)` by mass means 98 g `H_(2)SO_(4)` are PRESENT in 100 g of the solution `=100//1.84cm^(3)` `=54.35cm^(3)=0.5435L` Molar mass of `H_(2)SO_(4)="98 g mol"^(-1)` `therefore"MOLARITY of the GIVEN solution"("1 mol")/(0.5435L)=18.4M` Applying `""M_(1)V_(1)=M_(2)V_(2),18.4xxV_(1)=5xx0.500 or V_(1)=0.136L=136cm^(3)`

Discussion

45.	Concentrated aqueous sulphuric acid is 98% acid required to make one litre of 1.80 g mL^(-1). Volume of acid required to make one litre of 0.1 M H_(2)SO_(4) solution is :
Answer» 16.65 ML 22.90 mL 11.10 mL 5.55 mL Solution :Volume of 98% (W/W) solution of `H_(2)SO_(4)` `=("Mass")/("density")=((100g))/((1.80g mL^(-1))` =55.55 mL = 0.055L `"Molarity (M)"((98G))/((98g mol^(-1))xx(0.055L))` `=18.02 mol^(-1)L=18.02 M` `Vxx(18.02 M)=1Lxx(0.1M)` `V=((1L)xx(0.1M))/((18.02 M))=0.0055 L` =5.5 mL

Discussion

46.	Concentrated aqueous sulphuric acid is 98% H_(2)SO_(4) by mass and has a density of "1.80 g mL"^(-1). Volume of the acid required to make one litre of 0.1 M H_(2)SO_(4) solution is
Answer» 5.55 mL 11.10 mL 16.65 mL 22.20 mL Solution :`98% H_(2)SO_(4)` by mass means `98 g _(2)SO_(4)` PRESENT in 100 g of the solution, i.e., 100/1.80 mL = 55.5 mL Also `"98 g "H_(2)SO_(4)="1 mole "H_(2)SO_(4)` `therefore"Molarity of 98% "H_(2)SO_(4)` solution `=(1)/(55.5)xx"1000 mol L"^(-1)` Now, `M_(1)V_(1)=M_(2)V_(2).` HENCE, `(1000)/(55.5)MxxV_(1)="1000 mL"xx0.1M` `"or"V_(1)=5.55mL`

Discussion

47.	Concentrated aqueous sodium hydroxide can separate a mixture of:
Answer» `A1^(3+)` and `SN^(2+)` `A1^(3+)`and `FE^(3+)` `A1^(3+)`and`ZN^(2+)` `Zn^(2+)` and `Pb^(2+)` Answer :B

Discussion

48.	Concentrated aqueous sodium hydroxide can separate a mixture of
Answer» `AL^(3+)` and `Sn^(2+)` `Al^(3+)` and `Fe^(3+)` `Al^(3+)` and `Zn^(2+)` `Zn^(2+)` and `PB^(2+)` ANSWER :B

Discussion

49.	Conce. HNO_(3) on dehydration with P_(4)O_(10) gives .......... .
Answer» SOLUTION :`N_(2)O_(5)`

Discussion

50.	Concentated aqueous sodiumhydroxide cannot separate a mixture of
Answer» `Al^(3+)` and `SN^(2+)` `Al^(3+)` and `FE^(3+)` `Al^(3+)` and `ZN^(2+)` `Zn^(3+)` and `Pb^(2+)` Answer :a,C,d

Discussion

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Concentration of Ca ion in blood plasma is(approximately)

Concentration of Cl^(-) ions in a solution obtained by mixing 600 ml of 0.1 M CaXl_(2) and 400 ml of 0.4 M AlCl_(3).

Concetration of a reactant A is changed from 0.044 M to 0.032 M in 25 minutes, the average rate of the reactio during this interval is

Concentration described by percentage is commonly used in industrial, chemical industries.

Concentration is expressed in ?

concentrated sulphuric acid on reaction with NaCl, NaBr and Nal produces HCl, bromine and iodine respectively. What order of oxidising ability of halogens with reference to sulphuric acid can be established on the basis of this reaction ?

Concentrated sulphuric acid is approximately 18 molar. 5 cc of it are added to make 500 cc of the solution. The approxi mately normality of the solution will be

Concentrated sulphuric acid is added followed by heating of each of the following test tubes labelled (i) to Which of the following statement is incorrect about these observations ?

Concentrated sulphuric acid is added followed by heating in each of the following test types labelled (i) to (v) Identify in which of the above test tubes, the following changes will be observed. Support your answer with the help of a chemical equation. evolution of colourless gas

Concentrated sulphuric acid is added followed by heating in each of the following test types labelled (i) to (v) Identify in which of the above test tubes, the following changes will be observed. Support your answer with the help of a chemical equation. formation ofblack substance

Concentrated sulphuric acid is added followed by heating in each of the following test types labelled (i) to (v) Identify in which of the above test tubes, the following changes will be observed. Support your answer with the help of a chemical equation. evolution of brown gas

Concentrated sulphuric acid does not act as:

Concentrated sulphuric acid acts as

Concentrated nitric acid used in the laboratory work is 68% nitric acid by mass in aquequs solution. What should be molarity of such sample of the acid if the density of solution is 1.504 g mL""^(-1) ?

Concentrated nitric acid used in the laboratory work is 68% nitric acid by mass in aqueoussolution. What should be the molarity of such a sample of the acid if the density of solution is 1.504 g mL^(-1)?

Concentrated nitric acid used in the laboratory is 68% nitric acid by mass aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is "1.504 g mL"^(-1)?

Concentrated nitric acid used in laboratory work is 68% nitric add by aqueous solution. What should be the molarity of such a sample of the add lf the density of the solution is 1.504 g mL^(-1) ?

Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution 1.504 gm L^(-1) is ?

Concentrated nitric acid used as laboratory reagent is usually 69% by mass of nitric acid. Calculate the volume of the solution which contains "23 g HNO"_(3). (The density of the concentrated nitric acid solution is "1.41 g cm"^(-3)).

Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation of:

Concentrated nitric acid reacts with iodine to give:

Concentrated nitric acid can be transported in vessels of iron. Why ?

Concentrated nitric acid is added before proceeding to test for group III members. This is to

Concentrated nitric acid on heating decomposes to give:

Concentrated nitric acid can be transported in vessels of aluminium or tin cans. Why ?

Concentrated hydrochloric acid when kept in open air sometimes produces a cloud of white fumes. The explanation for it is that

Concentrated hydrochloric acid when kept in open air sometimes produces a could of white fumes. The explanation for it is that

Concentrated HCl is 10N,1000c c of 1N HCl can be obtained by diluting

Concentratged H_(2)SO_(4) is not used to prepare HBr from KBr becaue it

Concentrated H_(2)SO_(4)will not dehydrate the following complex :

Concentrated H_(2)SO_(4) will not dehydrate the following complex:

Concentrated H_2SO_4 is not used to prepare HBr from KBr because it:

Concentrated H_2SO_4 has great affinity for:

Concentrated aqueous sulphuric acid is 98% H_(2)SO_(4) by mass and has a density of 1.84"g cm"^(3). What volume of the concentrated acid is required to make 5.0 litre of 0.500 M H_(2)SO_(4) solution?

Concentrated aqueous sulphuric acid is 98% acid required to make one litre of 1.80 g mL^(-1). Volume of acid required to make one litre of 0.1 M H_(2)SO_(4) solution is :

Concentrated aqueous sulphuric acid is 98% H_(2)SO_(4) by mass and has a density of "1.80 g mL"^(-1). Volume of the acid required to make one litre of 0.1 M H_(2)SO_(4) solution is

Concentrated aqueous sodium hydroxide can separate a mixture of:

Concentrated aqueous sodium hydroxide can separate a mixture of

Conce. HNO_(3) on dehydration with P_(4)O_(10) gives .......... .

Concentated aqueous sodiumhydroxide cannot separate a mixture of