52671 + Interview Questions in Chemistry in Class 11

1.	The molarity of pure water is :
Answer» 18 M 50.0 M 55.6 M 100 M Solution :N//A

Discussion

2.	The molarity of a solution obtained by mixing 750 mL of 0.5 (M) HCl with 250 mL of 2(M)HCl will be:
Answer» 0.875 M 1.00 M 1.75 M 0.0975 M ANSWER :A

Discussion

3.	The molarity of 5.6V H_2O_2 is
Answer» 0.2 0.5 1 2 Answer :B

Discussion

4.	The Molarity of 200 ml of HCl solution which can neutralise 10.6 g. of anhydrous Na_(2)CO_(3) is
Answer» 0.1M 1 M 0.6 ml 0.75 M Answer :B

Discussion

5.	The molarity of 2 mole HCl in 5 lit aq. solution is ........
Answer» 10 2.5 0.4 4 Solution :MOLARITY `= ("MOLE")/(LITRE")= 2/5= 0.4M`

Discussion

6.	The molarity and % w/v of 40 volume H_2O_2 is ....and …..respectively.
Answer» 3.57 and 12.14 3.57 and 12.4 35.7 and 1.24 35.7 and 14.1 Solution :`M=V/11.2 =40/11.2`=3.57 M `M=(10xx % w//v)/34` % w/v = `(34xx3.57)/10`=12.14

Discussion

7.	The molarity and molality of a solution of sulphuric acid are 11.07 and 21.91 respectively. The density of the solution in g/ml is
Answer» Solution :Molality (m), MOLARITY(M), DENSITY (d) and gram MOLECULAR weight of solute (100 for heptance) are related as `m=(Mxx1000)/(1000d-MxxGMW)(or)21.3=(11.01xx1000)/(1000d-11.01xx100)` Density of solution` =1.618g cc^(-1)`

Discussion

8.	The molar volume of KCl and NaCl ar 37.46mL and 27.94mL respectively. The ratio of the unite cube edges of th crystals is
Answer» 1.296 1.116 1.341 0.95 Solution :(B) `(d_(KCL))/(d_(NACL))=((37.46)/(27.94))^(1/3)=1.116`

Discussion

9.	The molar volume of CO_(2) is maximum at
Answer» NTP `0^(@)C` and 2.0 atm `127^(@)C` and 1ATM `273^(@)C` and 2 atm Solution :(c ) From `n=(NRT)/(P)` At HIGH T andbelowp, V is high.

Discussion

10.	The molar volume of CO_2 is maximum at
Answer» NTP `0^@ Cand 2.0atm` `127^@Cand 1 atm` `8.3J MOL^(-1) K^(-1)` Answer :A

Discussion

11.	The molar volume of any gas at STP is
Answer» 1 litre 22.414 LIT `6.02xx10^(2` lit 22.414 ml ANSWER :B

Discussion

12.	The molar volume of an ideal gas at one atmosphere and 273^@C is
Answer» 22.4L 44.8L 11.2 L 5.6 L ANSWER :B

Discussion

13.	The molar volume of 22 g of CO_(2) is…..
Answer» Solution :MOLAR mass of `CO_(2)`=12 + 32=44g 44g of `CO_(2)` occupies molar volume=`2.24xx10^(-2)m^(3).` `:.` 2g of `CO_(2)` will occupy=`(2.24xx10^(-2))/cancel(44) XX cancel(22) =1.12xx10^(-2)L`

Discussion

14.	The molar solubility of Pbl_(2) in 0.2M Pb(NO_(3))_(2) solution in terms of solubility product, K_(sp)
Answer» `(K_(sp)//0.2) ^(1//2)` ` (K_(sp)//0.4) ^(1//2)` ` (K_(sp)//0.8 )^(1//2)` ` (K_(sp)//0.8 )^(1//3)` SOLUTION :` PbI_2 HARR Pb^(+2) +2I^(-) ` ` Pbunderset( 0.2 M) ((NO_3) ) _2to Punderset( 0.2M) b^(+2) + 2NO_3^(-) ` ` K_(sp)=(S+0.2) (2S)^(2) rArr S= (K_(sp)//0.8)^(1//2) `

Discussion

15.	The molar solubility of Cd (OH)_(2) " is " 1.84 xx 10^(-5) M. Calculate the expected solubility of Cd(OH)_(2)in a buffer solution of pH = 12.
Answer» SOLUTION :`Cd (OH)_(2) HARR underset(S)(Cd_((aq))^(2+)) + underset(10^(-2))( 2OH_((aq))^(-) )` `2.49 xx 10^(-14) = S(10^(-2))^(2) "" :. S = 2.49 xx 10^(10M) `

Discussion

16.	The molar solubility in mol. lit^(-1) of a sparingly soluble salt MX_(4), is S. The corresponding solubility product K_(sp), is given by the relation
Answer» `S= (K_(sp)//128 ) ^(1//4)` ` S= (218 K_(sp))^(1//4)` ` S= (256 K_(sp) )^(1//5)` ` S= (K_(sp) //256 )^(1//5) ` SOLUTION :` MX_4 hArr underset(S) M^(4+)+ 4underset( 4S) X^(-)` ` K_(sp) =[M^(4+ ) ][X^(-)] ^(4) = S( 4S)^(4) =256 S^(5) ` ` S = (K_(sp ) // 256) ^(1//5) `

Discussion

17.	The molar solubility (in mol L^(-1)) of a sparingly soluble salt MX_(4) is 'S'. The corresponding solubility product is 'K_(sp)'. 'S' in terms of 'K_(sp)' is given by the relation :
Answer» `S=((K_(SP))/(129))^(1//4)` `S=((K_(sp))/(256))^(1//5)` `S=(256 K_(sp))^(1//5)` `s=(128 K_(sp))^(1//4)` Solution :`{:(MX_(4),HARR,M^(4+),+,4X^(-)),(S,,S,,4S):}` `K_(sp)=[M^(4+)][X^(-)]^(4)=(S)(4S)^(4)=256S^(5)` or, `S=((K_(sp))/(256))^(1//5)`

Discussion

18.	The molar mass of Na_(2)SO_(4) is
Answer» SOLUTION :`Na_(2) SO_(4)` = MOLAR mass ` = (23xx2)+ (32xx1)+ (16xx4)` = 46 + 32 + 64= 142

Discussion

19.	The molar mass of an unkown organic liquid (M ~ 100) is determined by placing 5 mL of the liquid in a weighted 125mL conical flask with a piece of Al foil with a pin hole in it. The flask is heated in a Al foil with a pin hole in it. The flask is heated in a boiling water bath until the liquid evaporates to expel the air and fill the flsk with unknown vapour at atmospheric pressure. After cooling to vapour at atmospheric pressure. After cooling to room temperature the flask and its contents are room temperature the flask and its contents are reweighed. The uncertainty in which piece of apparatus causes the largest percentage error in the molar mass:
Answer» balance ` (+-0.01g)` BAROMETER ` (+-0.2mm HG)` FLASK` (+-1.0mL)` thermometer `(+-0.2^(@)C)` Answer :A

Discussion

20.	The molar mass of a solute X in g mol^(-1), if its 1% solution is isotonic with a 5% solution of cane sugar (molar mass = 342 g mol^(-)), is
Answer» iron CONTENT in haemoglobin is 0.35% by mass 1 mole of haemoglobin contains 56 g iron 1 mole of haemoglobin contains 224 g iron if iron content is increased to 0.56%, molar mass of haemoglobin would be HIGHER than 65000 g `mol^(-1)` SOLUTION :`%Fe=(4xx56)/(65000)xx100=0.35%` weight Fe in 1 mole = `4xx56=224g` if % Fe `uarr`, m. WT `uarr`

Discussion

21.	The molar mass of hydrogen is
Answer» `1.008g` `2.016g` `6.02xx10^(23)G` `2.016amu` ANSWER :B

Discussion

22.	The molar mass of a substance is 20g. The molecular mass of the substance is
Answer» 20g 20amu 10g 10amu Answer :B

Discussion

23.	The molar heat of formation of NH_(4)NO_(3)(s) is -367.54 kJand those of N_(2)O(g)and H_(2)O(l)are + 81.46kJand - 285.78 kJ respectively at 25^(@)Cand 1.0 atmospheric pressure. Calculate DeltaHand Delta U for the reaction NH_(4)NO_(3)(s) rarr N_(2)O(g) + 2H_(2)O(l)
Answer» SOLUTION :`DeltaH_("reaction") =Sigma Delta_(F) H (` Products ) `- Sigma Delta_(f) H(` Reactants) `Deltan_(g) = n_(p) - n_(R)= 1-0 = 1 ` ` Delta U = Delta H - Delta n_(g)RT`

Discussion

24.	The molar heat of formation of NH_(4)NO_(3)(g) is -367.54kJ and thoseof N_(2)O(g) and H_(2)O(l) are +81.46kJ and -285.78 kJ respectively at 25^(@)C and 1.0 atmosphericpressure. Calculate Delta H and DeltaU for the reaction.
Answer» Solution :`NH_(4)NO_(3)(s) rarrN_(2)O(g) + 2H_(2)O(l), DELTAH _("reaction") = SigmaDelta_(f) H (` Products `) -SigmaDelta_(f) H (` Reactants) `Deltan_(g) =n_(p)- n_(r) = 1-0 = 1` `DeltaU = DeltaH - Delta n_(g) RT`

Discussion

25.	The molar heat capacity of water at constant pressure, C, is 75 JK^(-1) mol^(-1). When 1.0KJ of heat is supplied to 100g of water which is free to expand, the increase in temperature of water is :
Answer» 1.2K 2.4K 4.8K 6.6K Solution :`Q= nC_(P) Delta T` `1000 = ((100)/(18)) (75) Delta T RARR Delta T rArr Delta T= 2.4`

Discussion

26.	The molar entropies of HI_((g)), H_((g)) and I_((g)) at 298K are 206.5, 114.6, and 180.7 J mol^(-1)K^(-1) respectively. Using the DeltaG^(@) given below, calculate the bond energy of HI. HI_((g)) rarr H_((g)) + I_((g)), DeltaG^(@) = 271.8kJ (Give your answer after divide with 49.7)
Answer» 282.4 298.3 290.1 315.4 Solution :`HI RARR H + I` `Delta S^(0) = S_(P) - S_(R ) = (114.6 + 180.7) - (206.5) = 88.8` `Delta G^(0) = Delta H^(0) - T Delta S^(0) = 271.8` `rArr Delta H^(0) = 271.8 + (298)/(1000) xx (88.8) = 298.3`

Discussion

27.	The molar heat capacity at constant volume of a system is 12.41J. mol^(-1). In an adiabatic expansion the temperature of one mole of that gas falls from 298K to 288K. Calculate the work done by the gas.
Answer» ANSWER :`-0.1241 KJ`

Discussion

28.	The molar entropies of HI(g), H(g) and I(g) at 298 K are 206.5, 114.6, and 180.7 J mol^(-1)K^(-1)respectively. Using the DeltaG^(@) given below, calculate the bond energy of HI.HI(g)rarrH(g)+I(g)," "DeltaG^(@)=271.8 kJ
Answer» `282.4 kJ "mol"^(-1)` `298.3 kJ "mol"^(-1)` `290.1 kJ "mol"^(-1)` `315.4 kJ "mol"^(-1)` Solution :`DeltaS^(@) =-206.5+114.6 +180.7 =88.8` `DeltaG^(@)=DELTAH^(@)-TDeltaS^(@)` `DeltaH^(@)=271.8 + 298 xx88.8 XX 10^(-3)` `DeltaH^(@)=298.3`

Discussion

29.	The molar enthalpy of vaporisation of acetoneis less than that of water. Why ?
Answer» SOLUTION :Enthalpy of vaporisation of water is more than that of acetone becausethere is STRONG hydrogen BONDING in `H_(2)O` MOLECULES.

Discussion

30.	The molar enthalpy of vaporisation of acetone is less than that of water. Why ?
Answer» Solution :Amount of heat REQUIRED to vaporise one mole of a liquid at constant tempeature and under standard pressure (1 BAR) is called its molar enthalpy of vaporisation `Delta_("vap") H^(Θ)`. Molar enthalpy of vaporisation of WATER is more than that of acetone because there is strong hydrogen bonding in `H_(2)O` molecule

Discussion

31.	The molar enthalpy of vapourisation of acetone is less than that of water. Why ?
Answer» Solution :Amount of heat required to VAPORISE one mole of a liquid at constant TEMPERATURE and under standard pressure (1 bar) is called its molar enthalpy of vaporisation `Delta_("vap") H^( THETA )`. Molar enthalpy of vaporisation of water is more than of acetone because there is STRONG hydrogen bonding in `H_(2) O` molecule.

Discussion

32.	The molar enthalpies of combustion of C_(2)H_(2)(g), C("graphite") and H_(2)(g) are - 310.62 kcal, -94.05 kcal and -68.32 kcal respectively. Calculate the standard enthalpy of formation of C_(2)H_(2)(g).
Answer» Solution :The given equations are : (i)`C_(2)H_(5)(g) + 5/2O_(2)(g) to 2CO_(2)(g) + H_(2)O(g)` `DELTAH = -310.62 kcal` (ii) `C("graphite") + O_(2) to CO_(2)(g)DeltaH = -94.05 kcal` (iii) `H_(2)(g) + 1/2O_(2)(g) to H_(2)O(g) DeltaH = -68.32 kcal` The required EQUATION is `2C("graphite") + H_(2)(g) to C_(2)H_(2)(g)DeltaH = ?` MULTIPLY eq.(ii) by 2 and add eqn (iii) and (iv) `2C("graphite") + 2O_(2)(g) to 2CO_(2)(g) DeltaH = -188.10 kcal` `H_(2)(g) + 1/2O_(2)(g) to H_(2)O(g)DeltaH = -68.32 kcal` (iv)`2C("graphite") + H_(2)(g) + 5/2O_(2)(g) to 2CO_(2)(g) + H_(2)O(g)` `DeltaH= -256.42kcal` Subtract eq.(i) `(i) C_(2)H_(2)(g) + 5/2 O_(2)(g) to 2CO_(2)(g) + H_(2)O(g)` `DeltaH = -310.62 kcal` Subtracting `2C("graphite") + H_(2)(g) to C_(2)H_(2)(g)` `DeltaH = 54.20 kcal`.

Discussion

33.	The molar concentration of the chloride ion in the solution obtained by mixing 300 mL of 3.0 M NaCl and 200 mL of 4.0 M solution of BaCl_(2) is :
Answer» 1.6 M 1.8 M 5.0 M 0.5 M Solution :N//A

Discussion

34.	The molality of the solution containing 45 g of glucose in 2 kg of water is ………….
Answer» Solution : `0.125m` `molality `= ("Number of moles of solute")/("MASS of the solvent in KG") = ((45)/(180))/(2) = (0.25)/(2) = 0.125m` (GLUCOSE = Molar mass = 180 g)

Discussion

35.	The molality of a solution containing 1.8g of glucose dissolved in 250g of water is ……………
Answer» ` 0 . 2 ` M ` 0 . 0 1 ` M `0 . 0 2`M `0 . 0 4 ` M ANSWER :D

Discussion

36.	The molality of a urea solution in which 0.0100 g of urea, [(NH_2)_2 CO] is added to 0.3000 dm_3 of water at S.T.P. is:
Answer» `5.55 XX 10^(-4) M` 33.3 M `3.33 xx 10^(-2) M` 0.555 M ANSWER :A

Discussion

37.	The molality of a solution of ethyl alcohol in water in 1.55 m. How many grams of ethyl alcohol are dissolved in 2kg of water ?
Answer» Solution :Molarity of solution `= (("MASS of "C_(2)H_(5)OH)/("Molar mass of "C_(2)H_(5)OH))/("Mass of solvent in kg")` `("1.55 mol kg"^(-1))=(W)/(("46 g mol"^(-1))XX(2kg))` `W=("1.55 mol kg"^(-1))xx("46 g mol"^(-1))xx(2kg)=142.6g`.

Discussion

38.	The molality of 2% (W/W) NaCl solution nearly
Answer» 0.02 m 0.35 m 0.25 m 0.45 m Answer :B

Discussion

39.	The molal elevation of an unknown solution is equal to the molal elevation of 0.1 M solution of urea. The concentration of unknown solution is :
Answer» 1M 0.01 M 0.1 M none of these Answer :C

Discussion

40.	The Mn^(3+) ion is unstable in solution and undergoes disproportionation to give Mn^(2+),MnO_(2),andH^(+) ion. Write a balanced ionic equation for the reaction.
Answer» SOLUTION :`Mn_((aq))^(+3)toMn_((aq))^(+2)+MnO_(2(s))+H_((aq))^(+)` HALF reaction … O.H.R. : `Mn_((aq))^(+3)toMnO_(2(s))` R.H.R. : `Mn_((aq))^(+3)toMn_((aq))^(+2)` Balancing the reaction on adding `e^(-)` O.H.R. : `Mn_((aq))^(+3)toMnO_(2(s))+e^(-)` R.H.R. : `Mn_((aq))^(+3)+e^(-)toMn_((aq))^(+2)` ADDITION of `H_(2)O` for the balancing of hydrogen and OXYGEN. O.H.R. : `Mn_((aq))^(+3)+2H_(2)OtoMnO_(2(s))+4H_((aq))^(+)+e^(-)` R.H.R. : `Mn_((aq))^(+3)+e^(-)toMn_((aq))^(+2)` Now on addition of O.H.R. and R.H.R. `2Mn_((aq))^(+3)+2H_(2)O_((L))toMnO_(2(s))+2Mn_((aq))^(+2)+4H_((aq))^(+)`

Discussion

41.	The moderator used in nuclear reactor is
Answer» `H_(2)O` `D_(2)O` Graphite Boron Solution :HEAVY water `(D_(2)O)`or graphite is GENERALLY USED as MODERATOR in nuclear REACTORS.

Discussion

42.	Mn^(3+) ions are unstable in solution and undergo disproportionation to give Mn^(2+), MnO_(2) and H^(+) ions. What will be the balanced equation for the reaction ?
Answer» Solution :The skeletal equation is `Mn^(3+)(aq)toMn^(2+)(aq)+MnO_(2)(s)toH^(+)(aq)` The equation can be BALANCED as follows. (i) Writing the oxidation numbers of all atoms, we have `overset(+3)(Mn^(3+))(aq)tooverset(+2)(Mn^(2+))(aq)+overset(+4-2)(MnO_(2))(s)+overset(+1)(H^(+))(aq)` Obviously, `Mn^(3+)` is simultaneously reducing to `Mn^(2+)` and OXIDISING to `MnO_(2)` (ii) THEREFORE, the half reactions corresponding to reduction and oxidation processes are as follows: `Mn^(3+)(aq)toMn^(2+)(aq)` (reduction half reaction) `Mn^(3+)(aq)toMnO_(2)(s)` (oxidatio half reaction ) (iii) Balancing of reduction half reaction : (a) The number of Mn atoms is the same on the two sides. (B) There is no O ATOM involved in the reaction. (c) Balancing charge by adding electron, we have `Mn^(3+)(aq)+e^(-)toMn^(2+)(aq)` (balanced reduction half reaction) (iv) Balancing of oxidation half reaction : (a) The number of Mn atoms is the same on the two sides. Since, the reaction proceeds in acidic medium, O atom can be balanced by the addition of two `H_(2)O` molecules on the left hand side. `Mn^(3+)(aq)+2H_(2)O(l)toMnO_(2)(s)` Balancing H atoms, we have `Mn^(3+)(aq)+2H_(2)O(l)toMnO_(2)(s)+4H^(+)` (c) Balancing charge by adding electrons, we have `Mn^(3+)(aq)+2H_(2)O(l)toMnO_(2)(s)+4H^(+)+e^(-)` (balanced oxidation half reaction) (d) Adding balanced half reactions, we get `{:(""Mn^(3+)(aq)+e^(-)toMn^(2+)(aq)),(Mn^(3+)(aq)+2H_(2)O(l)toMnO_(2)(s)+4H^(+)+e^(-)),(bar(2Mn^(3+)(aq)+2H_(2)O(l)toMn^(2+)(aq)+MnO_(2)(s)+4H^(+))):}` This is the balanced ionic equation for the given reaction.

Discussion

43.	The Mn^(3+) ion is unstable in solution and undergoes disproportionation to give Mn^(2+) , MnO_(2) and H^(+) ion. Write a balanced ionic equation for the reaction.
Answer» Solution :The skeletal equation is: `Mn_(aq)^(3+) RARR Mn_(aq)^(2+) + MnO_(2_((s))) + H_((aq))^(+)` Oxidation HALF equation: `Mn_((aq))^(3+) rarrMnO_(2_((s)))` Balance O.N. by adding electrons, `Mn_((aq))^(3+) rarrMnO_(2((s))) + e^(-)` Balance charge by adding `4H^(+)`ions Balance O atoms by adding `2H_(2)O` `Mn_((aq))^(3+)+2H_(2)OrarrMnO_(2((s)))+4H_((aq))^(+)+e^(-)` Balance O atoms by adding `2H_(2)O` `Mn_((aq))^(3+)+2H_(2)O_((l))rarrMnO_(2((s)))+4H_((aq))^(+)+e^(-)`.......................(1) Reduction half equation: `Mn^(3+)rarrMn^(2+)` Balance O.N. by adding electrons: ` Mn_((aq))^(3+) + e^(-) rarr Mn_((aq))^(2+)` ..............(2) Adding Equation (1) and (2), the balanced equation for the disproportionation reaction is `2Mn_((aq))^(3+)+2H_(2)O_((l)) rarr MnO_(2_((s)))+Mn_((aq))^(2+)+4H_((aq))^(+)`

Discussion

44.	The Mn^(3+) ion is unstable in solutin and undergores disproportionation to give Mn^(2+),MnO_(2) and H^(+) ion write a balanced ionic equation for the reaction
Answer» Solution :The skeletal equation is `Mn^(3+)+(aq)rarrMn^(2+)(aq)+MnO_(2)(s)+H^(+)(aq)` Oxidation half equation `overset(+3)Mn^(3+)(aq)rarr overset(3+)(aq)rarr MnO_(2)(s)+4H^(+)(aq)+E^(-)` balance O atoms by adding `2H_(2)O : overset(2+)Mn^(3+)rarr Mn^(2+)` balance O.N by adding ELECTRONS `Mn^(3+)(aq)^(-)rarrMn^(2+)(aq)` adding Eq (i) and Eq (ii) the balanced equation for the disportionation reaction is `2Mn^(3+)(aq)+2H_(2)O(l)rarrMnO_(2)(s)+4H^(+)(aq)+e^(-)` balance O atoms by adding `2H_(2)Mn(aq)+2H_(2)O(l)rarrMnO_(2)(s)+4H^(+)(aq)+e^(-)` reduction half equation `overset(+3)Mn^(3+)rarr overset(Mn^(2+)` ltbalance O.N by adding electron `Mn^(3+) (aq)+e^(-)rarrMn^(2+)(aq)` adding eq (i) and eq (ii) the balanced equation for disproportion reaction is `2 Mn^(3+)(aq)+2H_(2)O+2H_(2)O(l)rarrMnO_(2)(s)+Mn^(2+)(aq)+4H^(+)(aq)`

Discussion

45.	The mixtures are homogeneous irrespective of their physical state and such homogeneous mixtures are called as ___________
Answer» SOLVENT solute SOLUTIONS mixtures ANSWER :C

Discussion

46.	The mixture of exhanol and water are separated by ............. .
Answer» SOLUTION :AZEOTROPIC DISTILLATION

Discussion

47.	The mixture of hydrazine and hydrogen peroxide with copper (II) catalyst is used as a rocket propellant. Why ? Write the reactions involved.
Answer» Solution :When HYDRAZINE and hydrogen PEROXIDE reacrt in PRESENCE of `CU^(2+)` Catalyst the following reaction is takes place. `N_(2)H_(2)(I)+2H_(2)O_(2)(I)overset(Cu^(2+))toN_(2)(g)+4H_(2)O(g)` Since the formation of nitrogen and steam is accompaniedd with evolution of large amount of heat thus act as a rocked PROPELLANT.

Discussion

48.	The mixture of diethyl ether and ethanol can be purified by ............. .
Answer» SOLUTION :SIMPLE DISTILLATION

Discussion

49.	The mixture of BCl_(3) vapour and hydrogen gas is subjecteed to electric doscharge . The chief products are
Answer» B,HCl `B_(2)H_(6),HCl` `B_(2) H_(5)Cl,HCl` `B_(2)H_(4)CL_(2),HCl` Answer :2

Discussion

50.	The mixing or redistribution of energy among the atomic orbitals is known as hybridisation. In hybridisation each electron can be described by its wave function psi. Which of the following set of species has same electronic geometry
Answer» `PCl_(3), NH_(3),SO_(3) ` `CH_(4),NH_(3),H_(2)O` `CIF_(3),BF_(3),NF_(3)` `CO_(2),SiO_(2),SO_(2)` ANSWER :B

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

The molarity of pure water is :

The molarity of a solution obtained by mixing 750 mL of 0.5 (M) HCl with 250 mL of 2(M)HCl will be:

The molarity of 5.6V H_2O_2 is

The Molarity of 200 ml of HCl solution which can neutralise 10.6 g. of anhydrous Na_(2)CO_(3) is

The molarity of 2 mole HCl in 5 lit aq. solution is ........

The molarity and % w/v of 40 volume H_2O_2 is ....and …..respectively.

The molarity and molality of a solution of sulphuric acid are 11.07 and 21.91 respectively. The density of the solution in g/ml is

The molar volume of KCl and NaCl ar 37.46mL and 27.94mL respectively. The ratio of the unite cube edges of th crystals is

The molar volume of CO_(2) is maximum at

The molar volume of CO_2 is maximum at

The molar volume of any gas at STP is

The molar volume of an ideal gas at one atmosphere and 273^@C is

The molar volume of 22 g of CO_(2) is…..

The molar solubility of Pbl_(2) in 0.2M Pb(NO_(3))_(2) solution in terms of solubility product, K_(sp)

The molar solubility of Cd (OH)_(2) " is " 1.84 xx 10^(-5) M. Calculate the expected solubility of Cd(OH)_(2)in a buffer solution of pH = 12.

The molar solubility in mol. lit^(-1) of a sparingly soluble salt MX_(4), is S. The corresponding solubility product K_(sp), is given by the relation

The molar solubility (in mol L^(-1)) of a sparingly soluble salt MX_(4) is 'S'. The corresponding solubility product is 'K_(sp)'. 'S' in terms of 'K_(sp)' is given by the relation :

The molar mass of Na_(2)SO_(4) is

The molar mass of a solute X in g mol^(-1), if its 1% solution is isotonic with a 5% solution of cane sugar (molar mass = 342 g mol^(-)), is

The molar mass of hydrogen is

The molar mass of a substance is 20g. The molecular mass of the substance is

The molar heat of formation of NH_(4)NO_(3)(s) is -367.54 kJand those of N_(2)O(g)and H_(2)O(l)are + 81.46kJand - 285.78 kJ respectively at 25^(@)Cand 1.0 atmospheric pressure. Calculate DeltaHand Delta U for the reaction NH_(4)NO_(3)(s) rarr N_(2)O(g) + 2H_(2)O(l)

The molar heat of formation of NH_(4)NO_(3)(g) is -367.54kJ and thoseof N_(2)O(g) and H_(2)O(l) are +81.46kJ and -285.78 kJ respectively at 25^(@)C and 1.0 atmosphericpressure. Calculate Delta H and DeltaU for the reaction.

The molar heat capacity of water at constant pressure, C, is 75 JK^(-1) mol^(-1). When 1.0KJ of heat is supplied to 100g of water which is free to expand, the increase in temperature of water is :

The molar entropies of HI_((g)), H_((g)) and I_((g)) at 298K are 206.5, 114.6, and 180.7 J mol^(-1)K^(-1) respectively. Using the DeltaG^(@) given below, calculate the bond energy of HI. HI_((g)) rarr H_((g)) + I_((g)), DeltaG^(@) = 271.8kJ (Give your answer after divide with 49.7)

The molar heat capacity at constant volume of a system is 12.41J. mol^(-1). In an adiabatic expansion the temperature of one mole of that gas falls from 298K to 288K. Calculate the work done by the gas.

The molar entropies of HI(g), H(g) and I(g) at 298 K are 206.5, 114.6, and 180.7 J mol^(-1)K^(-1)respectively. Using the DeltaG^(@) given below, calculate the bond energy of HI.HI(g)rarrH(g)+I(g)," "DeltaG^(@)=271.8 kJ

The molar enthalpy of vaporisation of acetoneis less than that of water. Why ?

The molar enthalpy of vaporisation of acetone is less than that of water. Why ?

The molar enthalpy of vapourisation of acetone is less than that of water. Why ?

The molar enthalpies of combustion of C_(2)H_(2)(g), C("graphite") and H_(2)(g) are - 310.62 kcal, -94.05 kcal and -68.32 kcal respectively. Calculate the standard enthalpy of formation of C_(2)H_(2)(g).

The molar concentration of the chloride ion in the solution obtained by mixing 300 mL of 3.0 M NaCl and 200 mL of 4.0 M solution of BaCl_(2) is :

The molality of the solution containing 45 g of glucose in 2 kg of water is ………….

The molality of a solution containing 1.8g of glucose dissolved in 250g of water is ……………

The molality of a urea solution in which 0.0100 g of urea, [(NH_2)_2 CO] is added to 0.3000 dm_3 of water at S.T.P. is:

The molality of a solution of ethyl alcohol in water in 1.55 m. How many grams of ethyl alcohol are dissolved in 2kg of water ?

The molality of 2% (W/W) NaCl solution nearly

The molal elevation of an unknown solution is equal to the molal elevation of 0.1 M solution of urea. The concentration of unknown solution is :

The Mn^(3+) ion is unstable in solution and undergoes disproportionation to give Mn^(2+),MnO_(2),andH^(+) ion. Write a balanced ionic equation for the reaction.

The moderator used in nuclear reactor is

Mn^(3+) ions are unstable in solution and undergo disproportionation to give Mn^(2+), MnO_(2) and H^(+) ions. What will be the balanced equation for the reaction ?

The Mn^(3+) ion is unstable in solution and undergoes disproportionation to give Mn^(2+) , MnO_(2) and H^(+) ion. Write a balanced ionic equation for the reaction.

The Mn^(3+) ion is unstable in solutin and undergores disproportionation to give Mn^(2+),MnO_(2) and H^(+) ion write a balanced ionic equation for the reaction

The mixtures are homogeneous irrespective of their physical state and such homogeneous mixtures are called as ___________

The mixture of exhanol and water are separated by ............. .

The mixture of hydrazine and hydrogen peroxide with copper (II) catalyst is used as a rocket propellant. Why ? Write the reactions involved.

The mixture of diethyl ether and ethanol can be purified by ............. .

The mixture of BCl_(3) vapour and hydrogen gas is subjecteed to electric doscharge . The chief products are

The mixing or redistribution of energy among the atomic orbitals is known as hybridisation. In hybridisation each electron can be described by its wave function psi. Which of the following set of species has same electronic geometry